# A UV flux drop preceding the X-ray hard-to-soft state transition during the
2010 outburst of GX 339$-$4

Zhen Yan1 and Wenfei Yu1
1Key Laboratory for Research in Galaxies and Cosmology, Shanghai Astronomical
Observatory,Chinese Academy of Sciences,
80 Nandan Road, Shanghai 200030, China<EMAIL_ADDRESS>
(Accepted . Received … ; in original form )

###### Abstract

The black hole X-ray transient GX 339$-$4 was observed with the Swift
satellite across the hard-to-soft state transition during its 2010 outburst.
The ultraviolet (UV) flux measured with the filter UVW2 of the Swift/UVOT
started to decrease nearly 10 days before the drop in the hard X-ray flux when
the hard-to-soft state transition started. The UV flux $F_{\mathrm{UV}}$
correlated with the X-ray flux $F_{\mathrm{X}}$ as $F_{\mathrm{UV}}\propto
F_{\mathrm{X}}^{0.50\pm 0.04}$ before the drop in the UV flux. During the UV
drop lasting about 16 days, the X-ray flux in 0.4–10 keV was increasing. The
drop in the UV flux indicates that the jet started to quench 10 days before
the hard-to-soft state transition seen in X-rays, which is unexpected.

###### keywords:

accretion, accretion disks — black hole physics —ISM: jets and outflows—
X-rays:binaries — X-rays:individual (GX 339$-$4)

††pagerange: A UV flux drop preceding the X-ray hard-to-soft state transition
during the 2010 outburst of GX 339$-$4–References††pubyear: 2011

##  1 Introduction

Most known black hole (BH) X-ray binaries (XRB) are transient sources. They
usually stay in quiescence and occasionally undergo dramatic X-ray outbursts.
These BH transients are known to experience different X-ray spectral states
with particular X-ray spectral, timing and radio properties during an outburst
(McClintock & Remillard, 2006), including two major spectral states. One is
called the soft state, the X-ray spectrum of which is dominated by a multi-
color blackbody from an accretion disk (Shakura & Sunyaev, 1973). The other is
called the hard state, the X-ray spectrum of which is dominated by a powerlaw
with a photon index of $1.5<\Gamma<2.1$ (McClintock & Remillard, 2006). Radio
emission with a flat spectrum, which is thought as the self-absorbed
synchrotron emission from a compact steady jet, is also a characteristic of
the hard state (Fender, 2001). Suppressed radio emission in the soft state is
thought to indicate the jet is quenched (e.g. Fender et al., 1999). During the
transition from the hard state to the soft state, the X-ray spectral
properties are rather transitional and complex. This state is called the
intermediate state or the steep powerlaw state, the X-ray spectrum of which is
composed of a steep powerlaw component ($\Gamma\geq 2.4$) sometimes combined
with a comparable disk component (McClintock & Remillard, 2006). Many studies
attempted to explain spectral state transitions (e.g. Esin et al., 1997), but
to date there are still discrepancies between theories and observations
probably due to the limitation of the underlying stationary accretion
assumption, which is inconsistent with the observational evidence that non-
stationary accretion plays a dominating role in determination of the
luminosity threshold at which the hard-to-soft state transition occurs (Yu &
Yan, 2009).

In the hard state, the radio flux and X-ray flux, if not considering outliers,
display a correlation in a form of $L_{\mathrm{radio}}\propto
L_{\mathrm{X}}^{0.5-0.7}$(e.g. Corbel et al., 2000, 2003, 2008; Gallo et al.,
2003), indicating the coupling between the jet and the accretion flow. A
similar correlation also holds between the optical/near-infrared (NIR) and the
X-ray flux (Homan et al., 2005; Russell et al., 2006; Coriat et al., 2009),
which suggests that the optical/NIR emission has the same origin as that of
the radio emission, i.e. the optically thick part of the jet spectrum can
extend to NIR or even optical band (Corbel & Fender, 2002; Buxton & Bailyn,
2004; Homan et al., 2005; Russell et al., 2006). In contrast to the extensive
studies of the relations between the radio, optical/NIR emission and the X-ray
emission, there are much less studies about the ultraviolet (UV) emission in
different X-ray spectral states because of limited UV observations. Swift/UVOT
has three filters to cover the UV band. Simultaneous observations with the
UVOT and the XRT onboard Swift provide unprecedented opportunities for us to
study the UV emission in different X-ray spectral states.

GX 339$-$4 is one of the best-studied BH X-ray transients which undergoes
frequent outbursts in the past decades. Because its donor star is very faint
(Shahbaz et al., 2001), GX 339$-$4 during outbursts is therefore an excellent
target to investigate the UV/optical/NIR emission from the accretion and
ejection in the various X-ray spectral states in detail. In this paper, we
focus on reporting the results obtained from our proposed Swift observations
of GX 339$-$4 during the 2010 outburst. We found the UV flux decreased well
before the hard-to-soft state transition occurred.

##  2 Observations and data reduction

In the past two decades GX 339$-$4 have undergone rather frequent outbursts.
There is a unique correlation found between the peak flux in the hard X-ray
and the outburst waiting time in this source (Yu et al., 2007; Wu et al.,
2010). This empirical relation allowed us to predict the peak flux in the hard
X-ray in an on-going outburst if the time when the hard X-ray flux reaches its
peak can be determined (Wu et al., 2010). GX 339$-$4 entered a new outburst
since the beginning of 2010 (Yamaoka et al., 2010; Tomsick, 2010; Lewis et
al., 2010), and then experienced an extended flux plateau of about 50 days
which was unusual. The accretion of disk mass during the extended plateau
period was expected to cause the actual hard X-ray peak lower than the
prediction. However we were still able to predict the time of the hard-to-soft
state transition based on a simple linear extrapolation of the rising trend
and the relation predicted in Wu et al. (2010). We requested a series of
target of opportunity (TOO) observations with Swift spanned by about 20 days,
which indeed covered the hard-to-soft state transition. These observations,
together with other TOO observations immediately before or after (under
Targets IDs 30943,31687) were analysed, which covered the period from January
21, 2010 to June 19, 2010 (MJD 55217–55366).

Due to high count rates, all the XRT observations of GX 339$-$4 during this
period were automatically taken in the windowed timing (WT) mode which only
provided one-dimensional imaging. We processed all the initial event data with
the task xrtpipeline to generate cleaned event files by standard quality cuts
for the observations. There was no cleaned event data generated for the
observation 00030943009, so we excluded this observation in our XRT data
analysis. We extracted all the cleaned WT mode events using XSELECT v2.4 as
part of the HEASOFT v6.11 package. Source extraction was performed with a
rectangular region of 20 pixels $\times$ 40 pixels centered at the coordinate
of GX 339$-$4 and the background extraction was performed with two rectangular
regions of 20 pixels $\times$20 pixels whose centers are 50 pixels away from
the coordinate of GX 339$-$4\. Some observations with high count rates ($>$
150 c/s) are affected by the pile-up. The easiest way to avoid distortions
produced by the pile-up is to extract spectra by eliminating the events in the
core of the point spread function (PSF). The method by investigating the grade
distribution (Pagani et al., 2006; Mineo et al., 2006) was used to determine
the excluded region, which was chosen as the central 2 pixels$\times$20
pixels, 4 pixels$\times$20 pixels, 6 pixels$\times$20 pixels and 8
pixels$\times$20 pixels when the count rate was in the range of 150–200 c/s,
200–300 c/s, 300–400 c/s and 400–500 c/s, respectively. After events
selection, we produced the ancillary response files (ARFs) by the task
xrtmkarf. The process included inputting exposure maps with xrtexpomap and
applying the vignetting and PSF corrections. The standard redistribution
matrix file (RMF, swxwt0to2s6_20010101v014.rmf) in the CALDB database was used
for the spectral analysis. All the spectra were grouped to at least 20 counts
bin-1 using grppha to ensure valid results using $\chi^{2}$ statistical
analysis.

Swift/UVOT has six filters: V, B, U, UVW1, UVM2 and UVW2. We applied aspect
corrections to all the UVOT data by using uvotskycorr. For each observation,
the sky image of each filter was summed using uvotimsum in order to increase
photon statistics, then the spectral files which are compatible with XSPEC can
be created by uvot2pha. We performed aperture photometry with the summed
images by using uvotsource with an aperture radius 5″. The background region
was chosen in a larger void region, which is away from the target. The
photometry results in AB magnitude system are shown in the upper panel of Fig.
2.

##  3 Data analysis and results

Figure 1: From top to bottom, Swift/XRT spectral parameters column density
$N_{H}$, inner disk temperature $T_{in}$, photon index, reduced $\chi^{2}$,
the RXTE/ASM and Swift/BAT lightcurves, and the hardness ratio between the BAT
and the ASM fluxes. The vertical dotted and dashed lines indicate the time
interval when the source transited between the hard and the soft state.

All the XRT spectra were fitted in XSPEC v12.7.0. The photons with energies
below 0.4 keV were ignored during the spectral fits because of hot pixel
problems at lower energy (Kim Page, private communication). The 0.4–10 keV
X-ray spectra during the period of MJD 55217–55293 can be fitted well by a
model composed of an absorbed powerlaw with an photon index $\sim$ 1.61–1.79,
which indicates the source was in the hard state. Then the photon index
steepened ($\Gamma\sim$ 2.09–2.49) in the period of MJD 55297–55324 and an
additional disk component was required after MJD 55302, indicating the source
was leaving from the hard state to the soft state (see also Motta et al.,
2010). The fitted $N_{H}$ shows a sudden increase during the state transition.
A possible reason for the high absorption is that the powerlaw model does not
exhibit a low energy cut-off which is caused by the seed photon of the
Comptonization. We fitted the XRT spectra in order to get the photon index and
to estimate the X-ray flux, and the discussion of the physical spectral
components is beyond the scope of this paper. So we used the simple powerlaw
model rather than the Comptonization models for simplicity. The 0.4–10 keV
X-ray spectra taken after MJD 55334 can be fitted with only an absorbed multi-
color disk blackbody model ($T_{in}\sim$ 0.84–0.87 keV), indicating that the
source has entered the soft state. The best-fit spectral parameters are
plotted in Fig. 1. These model parameters were used to show the spectral
states the source resided in (see Fig. 1) and to estimate the 0.4–10 keV X-ray
flux (see Fig. 2). We also plotted the public daily lightcurves of GX 339$-$4
obtained from the RXTE/ASM and the Swift/BAT in Fig. 1. The X-ray spectral
states are distinguishable by the hardness ratio between the count rates in
the BAT (15–50 keV) and in the ASM (2–12 keV) (see also Yu & Yan, 2009; Tang
et al., 2011), consistent with the XRT spectral results (see Fig. 1).

Figure 2: The X-ray and UV/optical lightcurves of GX339$-$4 across the hard-
to-soft state transition during the 2010 outburst. Spectral state boundaries
are indicated by the vertical lines. The UVOT photometry results are plotted
in the upper panel. The UVW2 filter was used through all the observations. The
UV flux initially increased with the X-ray flux, and then decreased after MJD
55285, while the X-ray flux was still increasing.

The upper panel of Fig. 2 shows the magnitudes of six UVOT filters in
different X-ray spectral states. The UVW2 filter was used throughout all the
observations. The flux in the UVW2 band increased with the X-ray flux at the
beginning, and then began to decline about 10 days before the hard-to-soft
state transition started. The UVW2 flux decreased by about 62% in a period of
16 days since MJD 55285 and then kept at a steady level (see Fig. 2). Yale
SMARTS XRB project 111http://www.astro.yale.edu/buxton/GX339/ has offered
optical/NIR photometric coverage of this outburst on a daily basis. The
optical/NIR flux also showed a drop (see also Russell et al., 2010a; Cadolle
Bel et al., 2011; Buxton et al., 2012), which was simultaneous with the UV
flux drop within the uncertainties of the sampling. The flux decreased by
about 80% in V band, 84% in I band, 92% in J band and 94% in H band,
respectively. As we can see, the decreased amplitudes were wavelength
dependent, i.e. the flux decreased more at lower frequencies than at higher
ones (see also Homan et al., 2005). Moreover, such a drop in radio flux was
also seen (Zhang et al., 2012), implying a common physical origin for the flux
drops in the broad bands.

We further investigated how the UV flux correlated with the X-ray flux in the
hard state and across the hard-to-soft state transition during the 2010
outburst of GX 339$-$4 (see Fig. 3). The UV flux correlated with the
unabsorbed 0.4–10 keV X-ray flux in a power-law form with an index of $\sim
0.50\pm 0.04$ before its drop. The X-ray flux and the UV flux in the soft
state did not follow this correlation (see Fig. 3), indicating that the
origins of the UV emission differ between the hard state and the soft state.

Figure 3: UVW2 band flux vs. 0.4-10 keV X-ray flux. The circles represent the
data when the UV flux increased, the diamonds represent the data when the UV
flux decreased, the triangle represent the data in the soft state, and the
squares represent the rest of data. The data with maximal UV flux is the end
of UV flux increase and the start of the UV flux decrease, so two symbols (a
circle and a diamond) were plotted on this data. The solid line shows the best
power-law fit to the data during UV flux increasing (circle symbols) with an
index of $0.50\pm 0.04$. The dotted line shows how the X-ray and UV fluxes
evolve, and the arrow indicates the beginning.

##  4 Discussion

The origin of the UV emission of black hole X-ray binaries in the hard state
has not been well-understood. The UV emission could be from the jet, the cool
disk, the hot ADAF and X-ray reprocessing in the accretion disk (Markoff et
al., 2003; Yuan et al., 2005; Rykoff et al., 2007; Maitra et al., 2009). Let’s
look at these options in details. The correlation of the form of
$F_{\mathrm{UV}}\propto F_{\mathrm{X}}^{0.50\pm 0.04}$ (see Fig. 3) before the
UV flux drop is quantitatively inconsistent with a disk origin of the UV
emission, since the UV flux $F_{\mathrm{UV}}$ should be proportional to the
X-ray flux as $F_{\mathrm{X}}^{0.3}$ for a simple viscously heated disk (see
Russell et al., 2006). X-ray reprocessing in the accretion disk could be a
promising option if not considering the UV flux drop before the hard-to-soft
transition we report in the paper. van Paradijs & McClintock (1994) has
predicted the emission of X-ray reprocessing should be proportional to
$L_{\mathrm{X}}^{0.5}$, which was seen in XTE J1817$-$330 during the decay of
the 2006 outburst in which the UV luminosity was found to correlate with the
2–10 keV X-ray luminosity as $L_{X}^{\alpha}$ where $\alpha=0.47\pm 0.02$
(Rykoff et al., 2007). In the case of GX 339$-$4 the UV flux drop was
accompanied by an X-ray increase, against the X-ray reprocessing
interpretation. Therefore the origin of the UV emission before the UV flux
drop as X-ray reprocessing by the outer disk is also ruled out.

The flux drop in the radio or optical/NIR band usually associated with the
hard-to-soft state transition is thought due to the jet quench (Fender et al.,
1999; Homan et al., 2005; Coriat et al., 2009). In the observations we
analyzed, a drop of the UV flux started about 10 days before the transition.
The flux drop is similar to those drops in the radio, the NIR, and the
optical, which is highly suggestive of a common origin. So the drop in the UV
flux should be due to jet quenching as well.

After the jet quenched, the UV/optical/NIR fluxes declined to a steady flux
level, indicating additional components other than the compact jet contributed
to the UV/optical/NIR emission. The likely source is the X-ray irradiation,
since the steady UV/optical/NIR fluxes maintain through the soft state, and we
found the irradiation model diskir (Gierliński et al., 2009) can provide an
acceptable fit to the UVOT and XRT data in the soft state. Moreover when the
hard X-ray reached its maximum, the UV/optical emission was at a very low
level (see Fig. 2), indicating potential X-ray irradiation contributes little
UV/optical/NIR emission. Therefore, the dominant UV/optical/NIR emission
during the rising phase of the outburst was from the compact jet.

The scales of the flux densities corresponding to the drops in the
UV/optical/NIR bands (peak flux density minus the steady flux density after
the jet quenching) should be all contributed by the compact jet, if the X-ray
irradiation also contributed a steady level of UV/optical/NIR flux densities
during the jet quenching. Then the scales of the dereddened flux densities
corresponding to the drops in the UV/optical/NIR bands can be used to
construct a spectral energy distribution (SED) of the emission from the
compact jet. However, the observed SED in UV/optical/NIR bands is highly
influenced by the extinction, which is not well constrained in GX 339$-$4\. We
then used different color excesses to deredden the scales of the flux
densities corresponding to the drops in the UV/optical/NIR bands with the
extinction law of Cardelli et al. (1989). To convert the magnitudes obtained
from SMARTS to flux densities, we used the zero-point fluxes in Bessell et al.
(1998). We then constructed the SED in the wavelength from radio to
UV/optical/NIR in order to investigate the jet spectrum. The flux density of
UVW2 band reached its peak on MJD 55285, the closest radio observation (on MJD
55286) shows the radio flux densities were 22.38 mJy and 25.07 mJy at 5.5 GHz
and 9 GHz, respectively (Zhang et al., 2012). The two radio flux densities can
be well fitted by a power-law with an index $\sim$ 0.19, which is consistent
with the optically thick synchrotron emission of the jet. If using the color
excess of $E(B-V)=1.2$ given by Zdziarski et al. (1998), the extrapolation of
the radio spectrum coincides with the scale of the dereddened flux density
corresponding to the drop in the UVW2 band, indicating that the optically
thick spectrum of the jet might had extended to the UVW2 band. But the scales
of the dereddened flux densities corresponding to the drops in the V, I, J and
H bands are all below the extrapolation of the radio spectrum. We found the
UV-NIR SED can be well fitted with two power-law components. From the H band
to the I band, the power-law index was about $-0.27$, and from the I band to
the UVW2 band, the power-law index is about $+1.16$. The data in different
wavelengths were not exactly simultaneous. So this complex SED may be caused
by the large variation of the jet spectrum on the timescale of less than one
day in the optical/NIR band (Gandhi et al., 2011). If using the color excess
of $E(B-V)=0.94$ given by Homan et al. (2005), the scales of the dereddened
flux densities corresponding to the drops in the UV/optical/NIR bands were all
below the extrapolation of the radio spectrum and also show a turnover feature
in the I band. The turnover feature can be well fitted with two power-law
components with the power-law index $\sim-0.66$ from H band to I band, and
$\sim+0.16$ from the I band to the UVW2 band. If using the much smaller color
excess of $E(B-V)=0.66$ given by Motch et al. (1985), the scales of the
dereddened flux densities corresponding to the drop in the UV/optical/NIR
bands can be well fitted by a single power-law with an index of $-1.06$, which
is roughly consistent with the optically thin part of the jet spectrum.
Therefore, different color excess leads to different conclusion about the jet
spectral properties. The jet spectrum can not be well constrained in the
UV/optical/NIR bands due to large uncertainties in the estimation of the
extinction.

However, because there is a positive correlation between the UV and the X-ray
fluxes during the UV flux rise ($F_{\mathrm{UV}}\propto
F_{\mathrm{X}}^{0.50\pm 0.04}$, see also Fig. 3), we suggest that the
optically thick part of the jet spectrum could extend to the UVW2 band (see,
e.g., the discussion in Coriat et al., 2009). If this is true, the break
frequency between the optically thick and the optically thin parts of the jet
spectrum should have been in the UVW2 band or higher frequencies. Then a lower
limit of $-1.17$ on the spectral slope of the optically thin synchrotron
emission of the jet can be given by a power-law fit from the UVW2 band to 0.4
keV, which is steeper than the typical spectral slope ($\sim-0.7$) of the
optically thin synchrotron jet emission. Spectral slopes steeper than $-0.7$
have been found before, for example, $-0.95$ in GX 339$-$4 (Dinçer et al.,
2012) and $-1.5$ in XTE J1550$-$564 (Russell et al., 2010b).

If the UV flux is mainly from optically thick synchrotron emission of the
compact jet, the jet should be very energetic. We estimated the radiative flux
of the jet at the UV peak was $3.21\times
10^{-9}~{}\mathrm{ergs}~{}\mathrm{s}^{-1}~{}\mathrm{cm}^{2}$ by extrapolating
the radio spectrum from 5.5 GHz to UVW2 band. The corresponding 0.1-100 keV
X-ray flux, extrapolated from the 0.4–10 keV spectrum, was about $2.53\times
10^{-8}~{}\mathrm{ergs}~{}\mathrm{s}^{-1}~{}\mathrm{cm}^{2}$, corresponding to
$\sim 0.15~{}L_{\mathrm{Edd}}$ if adopting a distance of 8 kpc and a mass of
$10~{}M_{\sun}$ (Zdziarski et al., 2004). Taking reasonable radiative
efficiency of the jet of about 0.05 and the Doppler correction factor of about
1 (Fender, 2001; Corbel & Fender, 2002), the jet power $P_{\mathrm{jet}}$ is
about 2.5 times of the X-ray luminosity $L_{\mathrm{X}}$. This is larger than
previous estimates of $P_{\mathrm{jet}}/L_{\mathrm{X}}$ in the hard state of
GX 339$-$4, which were all less than 1 (e.g. Corbel & Fender, 2002; Dinçer et
al., 2012). Moreover, the jet power at the UV flux peak was greater than the
X-ray luminosity during a bright hard state of more than 0.1
$L_{\mathrm{Edd}}$, much higher than the critical value of $4\times
10^{-5}L_{\mathrm{Edd}}$ determined in Fender et al. (2003). It is worth
noting that a much lower jet power ( $P_{\mathrm{jet}}/L_{\mathrm{X}}\sim
0.05$) was seen in GX 339$-$4 at similar X-ray luminosity $\sim
0.12~{}L_{\mathrm{Edd}}$ (Corbel & Fender, 2002). Therefore similar X-ray
luminosity corresponds to a large range of jet power in the same source. This
phenomenon is similar to the hysteresis effect of spectral state transitions,
in which mass accretion rate is not the dominant parameter (Yu & Yan, 2009).
This suggests that the jet power might be determined by non-stationary
accretion, which leads to insignificant dependence on the mass accretion rate
or X-ray luminosity.

## acknowledgements

We would like to thank the anonymous referees for valuable comments and
suggestions. We thank Kim Page, Chris Done, Albert Kong and Wenda Zhang for
help on the Swift/XRT data analysis, Andrzej A. Zdziarski and Feng Yuan for
useful discussions and the Swift team for the scheduling the TOO observations.
We also acknowledge the use of data obtained through the High Energy
Astrophysics Science Archive Research Center Online Service, provided by the
NASA/Goddard Space Flight Center. This work is supported by the National
Natural Science Foundation of China (10773023, 10833002,11073043), the One
Hundred Talents project of the Chinese Academy of Sciences, the National Basic
Research Program of China (973 project 2009CB824800), and the starting funds
at the Shanghai Astronomical Observatory.

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# A Complete Congruence System for the Erdos-Straus Conjecture

Miguel Angel Lopez

###### Abstract

Abstract: In this paper we attack the Erdos-Straus conjecture by means of the
structure of its solutions, extending and improving the results of a previous
paper. Using previous results and supported by the works of Elsholtz and Tao
and Monks and Velingker we define a system of congruences for which there are
always solutions to the Erdos-Straus conjecture and which we conjecture to
include all prime numbers. For this purpose, and always taking into account a
result due to Mordell that limits the congruences admitting polynomial
identities to those that are not quadratic residues, we will adopt a
transversal approach and classify the solutions by their form and not by those
congruences that produce them. Thus we define two new types of solutions,
which we call Type A and B, and relate them to the already known Type II
solutions and study their properties. Finally we conjecture that every prime
number has at least one solution of Type A or B and we associate a congruence
and a general polynomial to each Type of solution.

email<EMAIL_ADDRESS><EMAIL_ADDRESS>
Keywords: Erdos-Straus, diophantine equation, unit fraction, egyptian
fraction, congruences, integral solution, prime numbers, quadratic residues

Mathematics Subject Classification: 11D72, 11A07, 11A41

##  1 Introduction

In 1948, Paul Erdös and Ernst G. Straus formulated a conjecture that states
the following: the equation

$\frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

has at least one solution where $x,$ $y,$ and $z$ are positive integers. There
are many modular identities that solve this equation, for example for the case
$n\equiv 2\pmod{3}$ we can use the following expression:

$\frac{4}{n}=\frac{1}{n}+\frac{1}{(n+1)/3}+\frac{1}{n(n+1)/3}$

L. J. Mordell studied this equation among many others in [1] and derived
identities for the cases $p\equiv 3\pmod{4}$, $2$ or $3\pmod{5}$, 5 or
$6\pmod{7}$ and $5\pmod{8}$. These combined identities solve all cases except
those where $p$ is congruent to 1, 121, 169, 289, 361 or $529\pmod{840}$.
Mordell himself wondered whether it might not be possible to find a sufficient
number of identities such that all possible cases would be completely covered.
This possibility was limited, however, by his discovery that if an identity
exists for a set of values $p\equiv r\pmod{q}$ then $r$ cannot be a quadratic
residue module $q$. For example, no such identity can exist for values of $p$
congruent to $1\pmod{q}$ since 1 is always a quadratic residue module $q$ for
any natural value of $q$. This implies, in fact, that it is not possible to
find a value $q$ such that identities can be found for all elements of
$\mathbb{Z}_{q}$.

The conjecture has been verified up to value $10^{7}$ by Yamamoto [2] and
$10^{14}$ by Swett [3]. Webb and others have shown that the natural density of
possible counterexamples to the conjecture is zero, for as $N$ tends to
infinity, the number of values in the interval $[1,N]$ that could be
counterexamples tends to zero.

In [4] we showed that, if $p=4k+1$, there exists a solution for $p$ such that

$\frac{4}{p}=\frac{1}{du}+\frac{1}{dv}+\frac{1}{duv}$

if and only if there exists $t\geq 0$ and a divisor $w$ of $k+1+t$ such that
$w\equiv-1\pmod{3+4t}$. In particular, if there exists a divisor $w$ of $k+1$
congruent to 2 module 3, $p$ has the previously mentioned solution. This
result is powerful even in the particular case where $t=0$ and in fact only by
employing this particular case all cases are ruled out except those where
$p\equiv 1\pmod{24}$ and also works in many other cases within this
congruence. That the problem cannot be solved completely with this result is
logical not only because, as Mordell rightly points out, 1 is quadratic
residue modulo 24, but also because this congruence contains the perfect
square of every prime number greater than 3, with repercussions that we will
explain later.

##  2 Main body

As in the previous article, we will establish a similar notation. We will
consider the values of each possible solution in an increasing order, i.e.
$x\leq y\leq z$. We will write a solution for each value $a\in\mathbb{N}$ as
what will allow us to list them in a more condensed form. We will use the
following definition:

######  Definition 1.

We say that a number $a\in\mathbb{N}$ is Egyptian of order 3 if there exists a
triplet of the form $(x,y,z)$ such that the Erdos-Straus conjecture is
fulfilled for the fraction $\frac{4}{a}$.

We will also use the following definition, used by Bradford in [5] and by
Elsholtz and Tao in [6]:

######  Definition 2.

We will say that a solution $(x,y,z)$ is of Type I if $gcd(a,x)=gcd(a,y)=1$
and $gcd(a,z)=a$ while a solution will be of Type II if $gcd(a,y)=gcd(a,z)=a$
and $gcd(a,x)=1$.

It is well known that prime numbers can only have these two types of
solutions. We will also point out that, contrary to the previous article, in
some cases we will consider the value a to be studied as a composite, and when
it is prime we will use the letter $p$. We will begin by recalling the result
of the previous article that motivates the new approach to the conjecture, and
it is the following.

######  Theorem 1.

Let be $p\in\mathbb{N}$ prime. Suppose that $p=4k+1$, there exists a solution
for $p$ with the form $(du,dv,duv)$ with $d,u,v\in\mathbb{N}$ if and only if
there exists $t\geq 0$ and a divisor $w$ of $k+1+t$ such that
$w\equiv-1\pmod{3+4t}$. In particular, if there exists a divisor $w$ of $k+1$
congruent to 2 module 3, $p$ is Egyptian of order 3 with a solution of the
form $(du,dv,duv)$.

We will not prove this result, for anyone interested in seeing its proof may
consult [4]. The proof is also constructive in an implicit way; following its
steps, it follows that

$d=\frac{k+1+t}{w},n=\frac{w+1}{3+4t},u=\frac{1+np}{4dn-1},v=np$

All these numbers are natural in a trivial way, without more than applying the
hypotheses, except in the case of $u$, for which it is simple to verify it by
substituting the values to arrive at $1+np\equiv 0\pmod{4dn-1}$. It can be
verified easily also that
$\frac{4}{p}=\frac{1}{du}+\frac{1}{dv}+\frac{1}{duv}$ because it is a mere
algebraic exercise.

It is also evident, without more than performing a modular arithmetic
calculation that, for the case $t=0$, if $k+1$ has a divisor $w$ congruent to
2 modulo 3, then either $w$ itself is a prime number or there exists another
divisor of $k+1$ congruent to 2 modulo 3 that is, so we can restrict our
search for valid $w$ divisors strictly to prime candidates.

As a result of the proof, in addition, the paper proved that for any value of
$t$ it is satisfied that $gcd(u,v)=1$. We can also come to a quick conclusion
as to whether it is one of the basic types of solution.

######  Theorem 2.

Let be $p\in\mathbb{N}$ prime of the form $p=4k+1$, if $p$ has a solution
$(du,dv,duv)$, satisfies that this solution is of Type II.

###### Proof.

Suppose without loss of generality that $u\leq v$. Monks and Velingker prove
in [7] that necessarily, since $du\leq dv$, $p$ does not divide $du$. Since we
know that

$\frac{4}{p}=\frac{1}{du}+\frac{1}{dv}+\frac{1}{duv}$

and therefore,

$4duv=p(1+u+v)$

we have then that $p\mid 4v$, and since $p$ is always odd, being of the form
$p=4k+1$, then $p\mid v$. This automatically implies that the solution
$(du,dv,duv)$ is of Type II. ∎

This reasoning can be replicated without variations for the case $p=4k+3$ and
it is trivial to observe that the condition is also fulfilled for the only
solution of the case $p=2$. The first limitation regarding this solution is
that it can never be assumed with these same properties for any value that
lacks Type II solutions, and such values, as Elsholtz and Tao already
commented, exist.

At this point, we will name these types of solutions.

######  Definition 3.

We say that a value $a\in\mathbb{N}$ has a solution of Type A if there exist
$d,u,v\in\mathbb{N}$ such that $a$ is Egyptian of order 3 with a solution
$(du,dv,duv)$. This solution is of Type II if $a$ is prime.

In the previous article it was shown that any prime number $p$ such that
$p\not\equiv 1\pmod{24}$ always has a solution of Type A. For this, it was
sufficient to use (1) and the particular case where $t=0$. It is possible to
keep digging in this direction until one finds huge lists of congruences with
Type A solutions associated, since it has not even been used in particular
cases other than the value of $t=0$. There are many papers that develop large
numbers of identities, but that approach has many limitations. It is
interesting to note that the congruence $p\equiv 1\pmod{24}$ includes the
values 25,49,121,169,289…All of them are perfect squares, which is no
coincidence.

######  Theorem 3.

Let $n$ be odd such that $n\geq 5$, $3\nmid n$.Then $n^{2}\equiv 1\pmod{24}$.
In particular, the square of every prime number greater than 3 belongs to this
congruence.

###### Proof.

We consider the value $n^{2}-1=(n-1)(n+1)$. Since $n$ is odd, we know that
$n\equiv 1,3\pmod{4}$. If $n\equiv 1\pmod{4}$ then $n+1\equiv 2\pmod{4}$ and
$n-1\equiv 0\pmod{4}$, which instantly implies that $n^{2}\equiv 1\pmod{8}$.
If $n\equiv 3\pmod{4}$ we reach the same conclusion by analogous reasoning. On
the other hand, since $3\nmid n$ we have that $n\equiv 1,2\pmod{3}$. If
$n\equiv 1\pmod{3}$ then $n-1\equiv 0\pmod{3}$ and therefore $n^{2}-1\equiv
0\pmod{3}$. Likewise, if $n\equiv 2\pmod{3}$ then $n+1\equiv 0\pmod{3}$ and
therefore $n^{2}-1\equiv 0\pmod{3}$. In both cases $n^{2}\equiv 1\pmod{3}$.
Thanks to the Chinese Remainder Theorem we finally have that $n^{2}\equiv
1\pmod{24}$. ∎

This result we have just proved extends an earlier classical one by Conway and
Guy published in [8] that said that the square of every odd number is
congruent to 1 modulo 8. The fact that this happens imposes limits on what can
be achieved with Type I and II solutions, as Schinzel and Yamamoto point out,
since they proved that the square of every natural number lacks both Type I
and Type II solutions. For a modern proof, see [6].

Solutions of Type A are transversal to the identity of prime and composite;
both types of numbers can possess a solution with such a structure. If a prime
number $p$ possesses a solution of the form $(du,dv,duv)$ then every multiple
$mp$ with $m>1$ possesses a solution of the form $(mdu,mdv,mduv)$ which still
possesses the structure $(Du,Dv,Duv)$ with $D=md$. The main difference lies in
the fact that, while the first solution is necessarily of Type II, the second
does not belong to that type since $gcd(mp,Du)\geq m$.

It is also immediate to verify that, if we try to expand (1) for composite
values $a$ of the form $a=4k+1$, then the implication from right to left is
immediately fulfilled (it is enough to take $d=\frac{k+1+t}{w}$,
$n=\frac{w+1}{3+4t}$, $u=\frac{1+np}{4dn-1}$, $v=np$, as we said before),
while the opposite implication is false, for which it is enough to look for
example at the value $a=25$, that has solutions of Type A but for which there
is no $t\geq 0$ such that $6+1+t$ has a divisor $w$ such that
$w\equiv-1\pmod{3+4t}$. These solutions can never be of Type II for this
value, being 25 a perfect square.

If a thorough analysis of prime numbers is performed, it can be seen that,
among the first 9000 natural numbers, the only prime numbers lacking Type A
solutions are 193 and 2521. In many articles these values suddenly appear as
numbers that offer various types of problems or resist classification, as for
example [9]. These two numbers, however, have a second common structure of the
form $d(uv,up,vp)$. Both numbers possess solutions of this type; 193 has e.g.

$(50,1930,4825)=5(10,2\cdot 193,5\cdot 193)$

and for 2521 we have the tern

$(638,55462,804199)=11(58,2\cdot 2521,29\cdot 2521)$

This will be our second structure to analyze.

######  Definition 4.

We say that a value $a\in\mathbb{N}$ has a solution of Type B if there exist
$d,u,v\in\mathbb{N}$ such that $a$ is Egyptian of order 3 with a solution
$(duv,dua,dva)$.

First, let us characterize in a simple way what a prime number must fulfill to
possess a solution of this type, and we will find an interesting parallelism
with the previous case.

######  Theorem 4.

Let be $p\in\mathbb{N}$ prime, let $d\in\mathbb{N}$. There exists a Type B
solution for $p$ if and only if it exists $n\in\mathbb{N}$ such that
$p\equiv-n\pmod{4dn-1}$. In addition this solution is of Type II.

###### Proof.

First we will prove the implication from left to right. We have a solution
that satisfies the equation

$\frac{4}{p}=\frac{1}{duv}+\frac{1}{dup}+\frac{1}{dvp}$

and which we can rewrite as

$4duv=p+u+v$

This implies that

$u=\frac{p+v}{4dv-1}$

and this number will only be natural if $4dv-1\mid p+v$ or, what is the same
thing, there exists a certain $n\in\mathbb{N}$ such that

$p+n\equiv 0\pmod{4dn-1}$

To prove the converse implication, let $n\in\mathbb{N}$ be such that the above
congruence is satisfied for a certain value $d\in\mathbb{N}$. We define
$u=\frac{p+n}{4dn-1}$, $v=n$. Both numbers are natural and it’s
straigthforward to prove that

$\frac{4}{p}=\frac{1}{duv}+\frac{1}{dup}+\frac{1}{dvp}$

Moreover, it always happens that $gcd(p,duv)=1$, otherwise $p^{2}$ would be
divisor of at least one of the three coordinates of the solution and such a
thing is impossible, as demonstrated by Monks and Velingker in [7]. That
classifies these solutions again in the Type II category, which was the last
thing we wanted to prove. ∎

An immediate and very interesting conclusion is that this characterization
does not use the fact that $p$ is a prime number, but it is necessary to
consider it to demonstrate that this solution is of Type II, since there are
composite numbers that have solutions of Type B but are not of Type II, for
example $n=6$. On the other hand there are also composite numbers that do not
have solutions of this type, for example $n=15$. As a joint conclusion, we
have that all the solutions of Type A and B are in particular of Type II when
they are referred to a prime number.

Again, Type B solutions are transversal to the identity of prime and
composite; both types of numbers can possess a solution with such a structure.
However, unlike the previous case, if a prime number $p$ has a solution of the
form $d(uv,up,vp)$ this does not imply that every multiple $mp$ with $m>1$ has
a solution of the form $dm(uv,up,vp)$.

As we have already mentioned above, perfect squares can never have solutions
of Type I or II. This implies, first of all, that if a perfect square $a$ has
a Type A solution, this solution cannot be of Type II and, therefore, must be
inherited from one of its divisors.

On the other hand, a stronger conclusion is the following:

######  Theorem 5.

A perfect square $s$ can never have a Type B solution,, whether or not it is
of Type II.

###### Proof.

If it does, then using (4) we have that there exist $d,n\in\mathbb{N}$ such
that

$s\equiv-n\pmod{4dn-1}$

and therefore, following the steps of the theorem, we construct the identity

$\frac{4}{s}=\frac{1}{dn\frac{s+n}{4dn-1}}+\frac{1}{sd\frac{s+n}{4dn-1}}+\frac{1}{sdn}$

This identity is of the form

$\frac{4}{s}=\frac{1}{P_{d,n}(s)}+\frac{1}{Q_{d,n}(s)}+\frac{1}{R_{d,n}(s)}$

where all these functions are polynomial in the variable $s$, and Mordell
proved that, then, $-n$ cannot be a quadratic residue module $4dn-1$, but it
is, since $s$ is a perfect square, and this is a contradiction. Therefore no
perfect square ever possesses a Type B solution. ∎

Another immediate conclusion is the following:

######  Corollary 1.

Let $d,n\in\mathbb{N}$, then $\left(\frac{-n}{4dn-1}\right)=-1$ with
$\left(\frac{a}{b}\right)$ equal to the Jacobi symbol. In particular if
$4dn-1$ is prime then $\left(\frac{n}{4dn-1}\right)=1$ and $n$ is always
quadratic residue module $4dn-1$ (but may or may not be if $4dn-1$ is a
composite number).

We can also characterize the Type B solutions for the particular case of prime
numbers $p=4k+1$.

######  Theorem 6.

Let be $p\in\mathbb{N}$ prime. Suppose that $p=4k+1$, there exists a solution
for $p$ with the form $(duv,dup,dvp)$ with $d,u,v\in\mathbb{N}$ if and only if
there exists $t\geq 0$ and two positive divisors $a,b$ of $k+1+t$ such that
$a+b=3+4t$.

###### Proof.

We suppose first that $p$ has a Type B solution, then exists
$d,n\in\mathbb{N}$ such that

$p\equiv-n\pmod{4dn-1}$

then

$4k\equiv-n-1\pmod{4dn-1}$

and we can deduce that

$k\equiv-dn^{2}-dn\pmod{4dn-1}$

This implies that exists $c\in\mathbb{Z}$ with $k+dn^{2}+dn=c(4dn-1)$. We can
also suppose that $c$ is natural because $4dn-1$ and
$k+dn^{2}+dn\in\mathbb{N}$. If we reorder the expression, we have that

$k+c=dn(4c-1-n)$

We define $c=t+1$, $t\geq 0$, so $k+1+t=dn(3+4t-n)$. We define $a=n$,
$b=3+4t-n$.We have trivially then that $a,b\in\mathbb{N}$, $ab\mid k+1+t$,
$a+b=3+4t$.

Conversely, lets suppose that $\exists a,b\in\mathbb{N}$, $t\geq 0$ such that
$ab\mid k+1+t$, $a+b=3+4t$. We can reverse all the process defining $a=n$,
$b=3+4t-a$. We define

$z=n(3+4t-n)=ab$

then

$k+1+t=dz,d\in\mathbb{N}$

and therefore

$k+1+t=3dn+4dnt-dn^{2}$

We define $t=c-1$, $c\geq 1$ because $t\geq 0$. Then $k+c=4dnc-dn-dn^{2}$ and
rearranging we arrive to $k+dn+dn^{2}=c(4dn-1)$, $c\geq 1$. It is
straightforward then that

$k\equiv-dn-dn^{2}\pmod{4dn-1}$

and this implies automatically that there are $d,n\in\mathbb{N}$ such that
$p\equiv-n\pmod{4dn-1}$ and therefore $p$ has a Type B solution. ∎

######  Corollary 2.

The theorem also holds if $p=4k+1$ but it’s not prime (we never use the
primality of $p$ in the demonstration).

In our previous paper, to obtain the characterization of (1), we employed the
following prior result, whose proof, similar to (4) but more complex in some
steps, can be read in [4].

######  Theorem 7.

Let be $p\in\mathbb{N}$ prime. There exists a Type A solution for $p$ if and
only if it exists $d,n\in\mathbb{N}$ such that $p\equiv-4d\pmod{4dn-1}$. In
addition the values $u$ and $v$ are coprime and the solution is of Type II.

Experimental evidence shows that every prime number always has at least one
solution with one of these two types of structure. This leads us to formulate
what is the main conjecture of this article:

######  Conjecture 1.

Let be $p\in\mathbb{N}$ prime, then exists $d,n\in\mathbb{N}$ such that
$p\equiv-4d\pmod{4dn-1}$ or $p\equiv-n\pmod{4dn-1}$. If this result is true,
then this congruence system covers all primes and the Erdos-Straus conjecture
is true.

It is very interesting to note that there is a modular relationship between
the type $-n$ and type $-4d$ values, since both sets of divisors are mutual
inverses in $\mathbb{Z}_{4dn-1}$.

The conjecture has been experimentally verified and is true for any prime
number less than or equal to 104729, that is, it has been verified for the
first 10000 prime numbers. Minor cases like $p=2$ or $p=3$ also satisfy the
conjecture. The number of solutions does not have a regular growth pattern: a
value as large as 83449 not only lacks Type B solutions, but also has only two
Type A solutions that are at the same time of Type II (and they are, since
83449 is a prime number, the only two of this type that it has, since it
cannot inherit others from any factor).

Both types of structure are necessary for the conjecture to be true: as
previously mentioned, 193 and 2521 do not have Type A solutions but do have
Type B solutions, and these are not isolated cases, since, for example, 66529
is another case. On the contrary, 23929 does not have Type B solutions but it
does have Type A solutions. Both types of solutions, therefore, complement
each other.

It may be asked whether it is possible to choose $d,n\in\mathbb{N}$ such that
$4dn-1$ is always a prime number. While such a thing is possible for a huge
number of cases of $p$, it is not possible to do so in general, and a
significant example is, again, the case $p=2521$. This number has only one
Type B solution which has two $4dn-1$ associated values, which are 87 and
1275, neither of which is a prime number. 2521 does, in fact, satisfy a
property that may perhaps explain in part why it is so elusive: all values
from 1 to 10 are quadratic residues in $\mathbb{Z}_{2521}$. In fact, all prime
numbers congruent to 1 module 840 fulfill it, precisely one of the congruences
that resisted Mordell when studying the conjecture and one of the most
resistant to being categorized, perhaps in part because of this special
property.

Elsholtz and Tao relate in [6] the Type II solutions to six-coordinate vectors
$(a,b,c,d,e,f)\in\mathbb{C}^{6}$ satisfying a system of equations. They call
the set of all these values $\Sigma_{n}^{\pi}$ and define an application
$\Pi_{n}^{\pi}$ between this set and the algebraic surface
$S_{n}=\\{{(x,y,z)\in\mathbb{C}^{3}:4xyz=nyz+nxz+nxy}\\}$ as follows:

$\Pi_{n}^{\pi}(a,b,c,d,e,f)=(abd,acdn,bcdn)$

These solutions are not just Type II, but $abd$ must be required to be coprime
with $n$. This condition becomes automatic when $n$ is a prime number, but is
not held in general.

It can be immediately observed that Type A solutions, with their values
ordered in any possible way, are given when $a=1$ or $b=1$ with the
identification $u=b$, $v=cn$ in the case $a=1$ and analogously when $b=1$.
When $c=1$ we obtain, precisely, the Type B solutions, performing this time
the automatic identification $u=a$, $v=b$. This fact suggests that these two
forms of solutions are canonically relevant, being obtained by substituting
for the smallest natural values for each of the three main parameters of the
Type II solutions. Also, if we allow $d=1$, we will obtain a very relevant
result, as we will see later.

The conjecture concerning the existence or non-existence of solutions of Type
A and B can be studied under a purely polynomial point of view, associating to
each type of solution an algebraic expression, as can be seen in the following
theorem.

######  Theorem 8.

Let $p$ be prime congruent to 1 modulo 4, we define the following polynomials
for $x,y,z,t\in\mathbb{N}$:

  * •

$P(x,y,t)=(4xy-1)(3+4t)-4x^{2}y$, $x,y,t\geq 0$

  * •

$Q(x,y,t)=(4xy-1)(3+4t)-4y$, $x,y,t\geq 0$

  * •

$R(x,y,t,z)=(4xyz-1)(3+4t)-4x^{2}y$, $x,y,z,t\geq 0$

  1. (i)

If $p=P(x,y,t)$ then $p$ has a Type B solution.

  2. (ii)

If $p=Q(x,y,t)$ then $p$ has a Type A solution.

  3. (iii)

If $p=R(x,y,t)$ then $p$ has a Type II solution.

All the implications, moreover, also hold conversely: if $p$ possesses a
solution of any of the above types, then it necessarily belongs to the image
of the corresponding polynomial.

###### Proof.

It is immediate to check, first, that the image of the three polynomials when
their variables take natural values are always numbers congruent to 1 modulo
4. Now, we start with (i). We know that

$p\equiv-x\pmod{4xy-1}$

which is equivalent by (4) to having a Type B solution. Now we consider
$p=4k+1$ with a Type B solution, by (6) we know that there exists $a,b$
divisors of $k+1+t$ such that $a+b=3+4t$. We rename $a=x$, $b=3+4t-x$, then
exists a certain $y$ such that

$k+1+t=xy(3+4t-x)$

and then, with elementary algebra,

$p=4k+1=(4xy-1)(3+4t)-4x^{2}y$

To prove (ii), if we define $p=(4xy-1)(3+4t)-4y$, it is immediate to prove
that

$p\equiv-4y\pmod{4xy-1}$

which by (7) implies that if $p$ is prime, then it has a Type A solution. To
prove the opposite, we consider that $p=4k+1$ has a solution of Type A, then
by (1) there exists a divisor $w$ of $k+1+t$ such that $w$ is congruent to
$-1$ modulus $3+4t$. This implies that $w+1=c(3+4t)$ for a certain $c\geq 1$,
and therefore there exists a certain natural number $e$ such that
$k+1+t=ew=e(c(3+4t)-1)$. Rearranging we arrive at

$k+1+t=ce(3+4t)-e$

which implies that

$p=4k+1=(4ce-1)(3+4t)-4e$

By renaming $x=c$, $y=e$,we already have it.

To prove (iii) we rely on a classic result of Mordell, which can be read for
example in [1], which says that a value $n$ has a Type II solution if and only
if there exist natural values $a,b,c,d$ such that they satisfy

$(4abcd-1)d=an+b$

By clearing we obtain that

$n=4bcd-\frac{d+b}{a}$

which will be a natural number if and only if there exists a certain $w$ such
that $d+b=aw$. By making the change of variable $d=aw-b$ we get that

$n=4abc(aw-b)-w$

and this number can only be congruent to 1 modulo 4 if $w$ is congruent to 3
modulo 4 or, what is the same, there exists a certain positive value $s$ for
which $w=3+4s$. By making a new change of variable we get that

$n=4abc(a(3+4s)-b)-(3+4s)$

which, renaming the variables and with some easy manipulations, leads us to
the polynomial $R(x,y,z,t)$ we were looking for. The whole procedure is
immediately reversible so that the implication works both ways. ∎

######  Remark 1.

It is immediate to check that

  * •

$R(x,y,t,1)=P(x,y,t)$

  * •

$R(1,y,t,x)=Q(x,y,t)$

######  Remark 2.

A p-value has a solution that is both of Type A and B in the particular case
in which $x=1$; in that case $P(x,y,t)=Q(x,y,t)$ and, therefore,
$0=P(x,y,t)-Q(x,y,t)=4y(x+1)(x-1)$, that can only happen when $x=1$. Below we
offer an alternative way of characterizing these solutions that are both of
Type A and B.

We mention now some general statements of Type A and B solutions concerning
bounds for $d$ and other simple properties.

######  Proposition 1.

Let $p$ be a prime number congruent to 1 module 4. If $p$ has a solution of
Type A then $d\leq\lfloor\frac{p+3}{8}\rfloor$. This bound is also optimal.

###### Proof.

By (1), we know that if $p$ has a solution of Type A then there exists $t>0$
and $w\equiv-1\pmod{3+4t}$ such that $w\mid k+1+t$ and, in fact,
$d=\frac{k+1+t}{w}$. We can write $w=-1+n(3+4t)$ with $n\geq 1$. Then
$d=\frac{k+1+t}{3n+4nt-1}$. Therefore,

$d=\frac{k+1+t}{3n+4nt-1}\leq\frac{k+1+t}{4t+2}\leq\frac{k+1}{2}=\frac{p+3}{8}$

and then we have that $d\leq\lfloor\frac{p+3}{8}\rfloor$.The coordinate is
also reached whenever $t=0$, $n=1$, what implies that $w=2$ and $2\mid k+1$,
and therefore k is odd. For example, it is reached for $p=13$, for this value
$k=3$ and $d=\frac{k+1}{2}=2=\frac{p+3}{8}=\lfloor\frac{p+3}{8}\rfloor$.
Therefore the bound is optimal. ∎

######  Proposition 2.

Under the conditions of Proposition 1, if $t$ is such that $k+1+t$ has a
divisor $w$ that satisfies that $w\equiv-1\pmod{3+4t}$ then
$t\in[{0,\lfloor\frac{k-1}{3}\rfloor}]$.

###### Proof.

Suppose that $t>\frac{k-1}{3}$, then $3t>k-1$ and therefore $3+4t>k+2+t$. This
then implies that $k+1+t\not\equiv-1\pmod{3+4t}$ and, in particular, any
divisor $w$ of $k+1+t$ is less than $3+4t$, so there is no divisor $w$ of
$k+1+t$ such that it is satisfied that $w\equiv-1\pmod{3+4t}$. We have
therefore proved the contrapositive of what we were looking for. ∎

######  Remark 3.

There are many examples for which $t$ takes the maximum value of the interval,
e.g. $k=4$. In fact, the bound is always optimal whenever $k\equiv 1\pmod{3}$,
because if we consider $k=3u+1$ with $u\geq 0$ we have that, if we take $t=u$,

$k+1+t=3u+1+1+u=2+4u$ $3+4t=3+4u$

and in this case we have that $k+1+t\equiv-1\pmod{3+4t}$ and therefore we can
take $w$ as $k+1+t$.

The propositions 3 and 4 shows that there are also parallels between Type A
and B solutions when calculating their coordinates for $d$.

######  Proposition 3.

Let $p$ be an odd prime number, if $p$ has a solution of Type B,
$(duv,dup,dvp)$ then $d\leq\lfloor\frac{p+3}{8}\rfloor$. This bound is also
optimal.

###### Proof.

We know that if $p$ has a Type B solution, then the equation $4duv=p+u+v$ is
satisfied and therefore $d=\frac{p}{4uv}+\frac{1}{4u}+\frac{1}{4v}$. We know
that there are multiple values where $u=1$ and $v=2$ or vice versa (by
symmetry, they give rise to the same solution), and for those values we have
that $d=\frac{p}{8}+\frac{1}{8}+\frac{1}{4}=\frac{p+3}{8}$. For any larger
values of $u$ and $v$ we have therefore that
$d\leq\lfloor\frac{p+3}{8}\rfloor$.

On the other hand, if $u=v=1$ then, using the construction of (4), we have
that $n=v=1$, $u=\frac{p+n}{4dn-1}=\frac{p+1}{4d-1}=1$ and therefore
$4d-1=p+1$. But this then implies that $p$ is even, and by hypothesis $p$ is
always odd, so the above bound is correct. ∎

######  Proposition 4.

Under the conditions of Proposition 3, if there exists $t\geq 0$ and two
positive divisors $a,b$ of $k+1+t$ such that $a+b=3+4t$ then
$t\in[{0,\lfloor\frac{k-1}{3}\rfloor}]$.

###### Proof.

It’s obvious to prove that, given $n\in\mathbb{N}$, $max\\{a+b/ab\mid
n\\}=n+1$. Therefore there can’t be such divisors for $k+1+t$ when
$max\\{a+b/ab\mid k+1+t\\}<3+4t$, and this implies that $(k+1+t)+1<3+4t$,
which implies that $3+4t>k+2+t$, the same values for which we couldn’t find a
Type A solution, so again $t$ must belong to the interval
$[{0,\lfloor\frac{k-1}{3}\rfloor}]$. ∎

######  Proposition 5.

Let $a$ be any natural number, then if $a\equiv-n\pmod{4dn-1}$ we also have
that
$a\equiv-\left(\frac{a+n}{4dn-1}\right)\pmod{\left(\frac{4ad+1}{4dn-1}\right)}$
and the converse is also true. Both congruences lead us, moreover, to the same
Type B solution for a.

###### Proof.

In the first case, using (4), we arrive at a Type B solution of the form
$(duv,dua,dva)$ with $u=\frac{a+n}{4dn-1}$, $v=n$. In the second case we have
that $4d\left(\frac{a+n}{4dn-1}\right)-1=\frac{4ad+1}{4dn-1}$, so using the
same theorem we arrive at another Type B solution of the form $(duv,dua,dva)$,
with $v=\frac{a+n}{4dn-1}$ and $u=n$, because

$u=\frac{a+v}{4dv-1}=\frac{a+\frac{a+n}{4dn-1}}{4d\frac{a+n}{4dn-1}-1}=\frac{4adn-a+a+n}{4dn-1}:\frac{4ad+1}{4dn-1}=\frac{4adn+n}{4dn+1}=n$

Both solutions are the same except for a translation of their coordinates, and
therefore both congruences must be true or false simultaneously. The
operations also show that they are cyclic, and by means of one congruence the
other can be reconstructed and vice versa. ∎

This last proposition implies, therefore, that every Type B solution always
has at least two congruences associated with it, as we already saw in the case
of 2521, which possessed a single Type B solution but two associated
congruences, which are $2521\equiv-2\pmod{87}$ and $2521\equiv-29\pmod{1275}$.
It is a result that can be applied outside the context of the conjecture as a
proposition in the field of modular arithmetic.

######  Proposition 6.

Let $p$ be prime, then $p$ possesses a solution that is both of Type A and B
if and only if $p\equiv-1\pmod{4d-1}$ for some $d\in\mathbb{N}$.

###### Proof.

If $p\equiv-1\pmod{4d-1}$ then, by (4), it has a Type B solution. Moreover, as
$-4d\equiv-1\pmod{4d-1}$ by (5) that same congruence also generates a Type A
solution, which is the one we already had.

Now suppose that p has a solution that is both of Type A and B, i.e.,

$(x,y,z)=(du,dv,duv)=(DUV,DUp,DVp)$

with $d,u,v,D,U,V\in\mathbb{N}$. We can consider without loss of generality
that $u<v$. From the Type A structure of the solution we obtain that $xy=dz$,
and from the Type B structure we obtain that $yz=Dxp^{2}$. This implies that
$y^{2}=Ddp^{2}$. Substituting into the Type B solution we have that

$D^{2}U^{2}p^{2}=Ddp^{2}$

and therefore $DU^{2}=d$, which implies that $U^{2}=\frac{d}{D}$ and, in
particular, that $D\mid d$. Substituting $y^{2}=Ddp^{2}$ into the Type A
structure we obtain that

$dv^{2}=Dp^{2}$

Since $p\nmid u$ we know that $p\mid v$ and that implies that
$\frac{D}{d}=\frac{v^{2}}{p^{2}}\in\mathbb{N}$ and $d\mid D$. We conclude that
$d=D$ and this automatically implies that $v=p$ and $U=1$. This implies that
the solution has the form $(du,dp,dup)$. Since it is of Type B, we have, by
(4), that necessarily $p\equiv-1\pmod{4d-1}$. ∎

##  3 Appendix I: Solutions of Type C?

As was said in Remark 1 it happens that, if we have the polynomial
$R(x,y,t,z)=(4xyz-1)(3+4t)-4x^{2}y$ that has as its image all the values that
possess a Type II solution, the Type A and B solutions are related to it since
the Type A solutions are obtained in the particular case in which $x=1$ while
the Type B solutions are obtained when $z=1$. This motivates the idea of
considering what happens in the case in which $y=1$, for which we obtain the
polynomial $R(x,1,t,z)=(4xz-1)(3+4t)-4x^{2}$. We will study it in the
following theorem that encompasses this new type of solutions.

######  Theorem 9.

Let $n\in\mathbb{N}$ be of the form $n=4k+1$. They are equivalent:

  1. (i)

There exists $(x_{0},z_{0},t_{0})$ such that $n=R(x_{0},1,t_{0},z_{0})$;

  2. (ii)

There exist $d,m\in\mathbb{N}$ such that $-n\equiv 4d^{2}\pmod{4dm-1}$;

  3. (iii)

n has a solution of the Erdos-Straus conjecture with the form $(uv,uwn,vwn)$,
with $u,v,w\in\mathbb{N}$;

  4. (iv)

For a certain $t\geq 0$ exists $a,b\in\mathbb{N}$ such that $ab=k+1+t$,
$3+4t\mid a+b$.

We will consider this new solution as a Type C solution.

###### Proof.

First we will see that (i) implies (ii). We know that $n$ can be written as
$n=(4x_{0}z_{0}-1)(3+4t_{0})-4x_{0}^{2}$, then it is automatically satisfied
that

$n\equiv-4x_{0}^{2}\pmod{4x_{0}z_{0}-1}$

and so we have (ii) automatically by identifying $d=x_{0}$, $m=z_{0}$.

To show that (ii) implies (iii) we know that $-n\equiv 4d^{2}\pmod{4dm-1}$ and
therefore $-nm\equiv d\pmod{4dm-1}$. This implies that $d+nm=v(4dm-1)$ for a
certain $v\in\mathbb{N}$, so we have that $4dmv=d+nm+v$. Dividing by $dmvn$ we
have that

$\frac{4}{n}=\frac{1}{mvn}+\frac{1}{dv}+\frac{1}{dmn}$

which is of the form requested with $u=d$, $w=m$.

Now we start from (iii) and assume that we have a solution of the form
$(uv,uwn,vwn)$, so multiplying by $dmvn$ we have again that $4uwv=u+nw+v$.
This implies that $v(4uw-1)=u+nw$ and therefore that $nw\equiv-u\pmod{4uw-1}$.
If we substitute $n=4k+1$ this leads us to that

$4wk\equiv-u-v\pmod{4uw-1}$

and therefore that

$k\equiv-u^{2}-uw\pmod{4uw-1}$

Translating the congruence to an equality leads to the fact that
$k+u^{2}+uw=d(4uw-1)$ for a certain $d\in\mathbb{N}$ and hence

$k+d=4duw-uw-u^{2}$

which can be rewritten as

$k+d=u(w(4d-1)-u)$

Since we know that $d$ is a natural number we can make the change of variable
$d=t+1$, $t\geq 0$ and we have then that the equality can be written as

$k+1+t=u(w(3+4t)-u)$

This implies that $k+1+t$ can be decomposed as the product of two divisors
$a=u$, $b=w(3+4t)-u$, such that $a+b=w(3+4t)$ and therefore $3+4t\mid a+b$, as
we were looking for.

To prove finally that (iv) implies (i), we assume that we have two divisors
$a,b$ of $k+1+t$ for a certain $t\geq 0$ such that $3+4t\mid a+b$. Undoing the
same math as above we have that

$k+1+t=u(w(3+4t)-u)$

and therefore that

$k+1+t=3uw+4uwt-u^{2}$

This leads us to the fact that

$4k+4+4t=12uw+16uwt-4u^{2}$

and therefore

$4k+1=12uw+16uwt-(3+4t)-4u^{2}$

which by rearranging the terms leaves us with the expression

$n=(4uw-1)(3+4t)-4u^{2}$

which shows that $n=R(u,1,t,w)$ and therefore that $n$ belongs to the image of
the reduced polynomial $n=R(x,1,t,z)$, which was just what was said in section
(i). ∎

######  Corollary 3.

Given the nature as a quadratic equation of the congruence $-n\equiv
4d^{2}\pmod{4dm-1}$ and taking into account the notation of Legendre’s symbol
and that it always happens that

$\left(\frac{-1}{4u-1}\right)=-1$

we have then that a necessary condition for a number $n$ of the form $n=4k+1$
to have a solution of Type C will be that there exists a certain $u$ for which
$n$ is not a quadratic residue in $\mathbb{Z}_{4u-1}$. This will imply that
$-n$ will possess not only one but two solutions to the equation $-n\equiv
x^{2}\pmod{4u-1}$. If we denote by $x_{0}$ and $4u-1-x_{0}$ those solutions we
have that exactly one of them will be even and the other odd and, therefore,
either $\frac{x_{0}}{2}$ or $\frac{4u-1-x_{0}}{2}$ will be a natural number.
Then it will be an indispensable condition, for a solution of Type C to exist,
that this number be a divisor of $u$.

Although this type of solution does not seem necessary in experimental terms
to cover all the prime values congruent to 1 modulo 4, it can be added to the
conjecture we already have, by extending the system of congruences without any
cost:

######  Conjecture 2.

Let be $p\in\mathbb{N}$ prime, then exists $d,n\in\mathbb{N}$ such that
$p\equiv-4d\pmod{4dn-1}$, $p\equiv-n\pmod{4dn-1}$ or
$p\equiv-4d^{2}\pmod{4dn-1}$. If this result is true, then this congruence
system covers all primes and the Erdos-Straus conjecture is true.

Another immediate implication, thanks to the fact that (4) is true for all
natural numbers, not just prime numbers, is that there exist infinite pairs of
values of the form $4k+1$ such that their distance is equal to $4s$, with
$s\geq 1$ chosen at our option and satisfying the conjecture. Indeed, if we
call

$S(x,y,t)=R(x,1,t,y)=(4xy-1)(3+4t)-4x^{2}$

then we have that $S(x,y,t)-P(x,y,t)=4x^{2}(y-1)$, and given $s\in\mathbb{N}$,
if we define $n=S(1,s+1,t)$ and $n^{\prime}=P(1,s+1,t)$ then it is immediate
to check that $n-n^{\prime}=4s$ for every value $t\geq 0$ and we have made the
infinite pairs requested, which satisfy the Erdos-Straus conjecture by being
image of the polynomials mentioned above.

##  4 Appendix II: Reduction to Egyptian Numbers of Order 2

Although the values that have the form $n=4k+1$ turn out to be the most
problematic to study, we can also take advantage of their particular structure
to reach results that only they can fulfill, like this theorem that proves the
existence of infinite consecutive values that satisfy the Erdos-Straus
Conjecture, something that Manuel Bello Hernández, Manuel Benito and Emilio
Fernández already proved in [10], but that is shown here in a much simpler and
direct way and by means of an ingenious transformation.

######  Theorem 10.

Given $n\in\mathbb{N}$, there exists a chain of $n$ consecutive natural
numbers that satisfies the Erdos-Straus conjecture.

###### Proof.

Since identities are known for even values and of the form $n=4k+3$, it will
suffice to prove this for those of the form $n=4k+1$. First we will show that,
if there exists $d\in\mathbb{N}$ such that $\frac{4d-1}{k+d}$ is Egyptian of
order 2, then $n=4k+1$ satisfies the Erdos-Straus conjecture.

If $\frac{4d-1}{k+d}=\frac{1}{y}+\frac{1}{z}$ then
$\frac{4d}{k+d}=\frac{1}{k+d}+\frac{1}{y}+\frac{1}{z}$. We call $k+d=x$, and
therefore $d=x-k$ and it follows that

$\frac{4(x-k)}{x}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

which implies that

$4=\frac{4k+1}{x}+\frac{1}{y}+\frac{1}{z}$

We call $n=4k+1$ and we have that

$\frac{4}{n}=\frac{1}{x}+\frac{1}{ny}+\frac{1}{nz}$

which was what we were looking for.

Now, it is well known that $\frac{a}{b}$ is Egyptian of order 2 if and only if
there exist $u,v$ divisors of $b$ such that $a$ divides to $u+v$. Applied to
$\frac{4d-1}{k+d}$ we have that there exist $u,v$ divisors of $k+d$ such that
$4d-1$ divides to $u+v$. Note that if $k$ is odd it is automatically fulfilled
by taking $d=u=1$, $v=2$.

We consider $k$ of the form $n!-f(n)$, and take $d=f(n)$. For this form to be
satisfactory there must exist $u,v$ divisors of $n!$ such that $4f(n)-1\mid
u+v$. We look for $f(n)$ to be as large as possible, and that is the case if
it is also $u+v$, with $u\neq v$. We take therefore $u+v=n+(n-1)$. Equaling we
have that

$4f(n)-1=2n-1$

And therefore these values $u,v$ will always be possible and valid if
$f(n)\leq\lfloor\frac{n}{2}\rfloor$, which implies that all values $4k+1$,
with $k$ belonging to the interval $[{n!-\lfloor\frac{n}{2}\rfloor,n!-1}]$,
satisfy the Erdos-Straus conjecture. Since we can take $n$ as large as we want
and the length of the interval increases with $n$, we can find consecutive
values to our liking that satisfy this conjecture. ∎

######  Corollary 4.

Proving that there exists $d\in\mathbb{N}$ such that $\frac{4d-1}{k+d}$ is
Egyptian of order 2 is equivalent to finding a Type II solution for $4k+1$.

###### Proof.

If $\frac{4d-1}{k+d}=\frac{1}{y}+\frac{1}{z}$, then there exists $u,v\mid
k+d$, $4d-1\mid u+v$, and therefore $k+d=uvw$, $u+v=t(4d-1)$ for some certain
naturals $w,t$. By elementary calculations and calling $n=4k+1$ we obtain that

$n=4uvw-\frac{u+v}{t}$

Renaming $u=ct-v$, we have that $n=c(4vwt-1)-4v^{2}w$. By changing the names
of the variables, $n=(4xyz-1)c-4x^{2}y$. Since $n$ is congruent with 1 modulo
4, then $c$ is congruent with 3 modulo 4, and so $c=3+4s$, $s\geq 0$, and
$n=R(x,y,z,s)$. By (8), $n$ is then a value with a Type II solution. ∎

##  5 Conclusion

We have defined two types of solutions, Type A and B, which have allowed us to
propose a system of congruences that we conjecture gives a solution of the
Erdos-Straus conjecture to all prime numbers. This is one of the most
reasonable techniques for solving the Erdos-Straus conjecture: to obtain a
system of congruences such that they all have an associated polynomial
identity and that they completely cover all prime numbers. The first part,
that every congruence has an associated polynomial identity and therefore none
of them contain quadratic residues, has already been shown. It remains to be
seen whether counterexamples to this new conjecture are found, which is the
same as finding some prime number that lacks solutions of both Type A and Type
B and, if not found and therefore suspecting that the conjecture is true,
whether or not we have created an alternative formulation of the same problem
as difficult to prove or more difficult to prove than the original problem. So
far we have experimentally proved that our proposed system of congruences
gives a solution to the first ten thousand prime numbers without exceptions,
which gives us hope for its usefulness as a tool to solve the Erdos-Straus
conjecture.

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