Erdős-Straus conjecture... It was necessary to write the solution in a more General form:

\[ \frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \]

Decomposing on the factors as follows: 𝑝²−𝑠²=(𝑝−𝑠)(𝑝+𝑠)=2𝑞𝐿

\[ x=\frac{p(p-s)}{tL-q} \]

\[ y=\frac{p(p+s)}{tL-q} \]

𝑧=𝐿

Decomposing on the factors as follows: 𝑝²−𝑠²=(𝑝−𝑠)(𝑝+𝑠)=𝑞𝐿

\[ x=\frac{2p(p-s)}{tL-q} \]

\[ y=\frac{2p(p+s)}{tL-q} \]

𝑧=𝐿

#Diophantine #equation #number-theory #holistic_algebra #diophantine

𝐴²+𝐴𝐵+𝐵²=𝐶²
𝐹²+𝐹𝑍+𝑍²=𝐶²

Such a system is solved as standard. First, we write down the parametrization of one equation.

𝐴=𝑝²−𝑠²
𝐵=𝑠(𝑠+2𝑝)
𝐶=𝑝²+𝑝𝑠+𝑠²=𝑥²+𝑥𝑦+𝑦²
𝐹=𝑥²−𝑦²
𝑍=𝑦(𝑦+2𝑥)

And then we find the parameterization for the necessary parameters.

𝑝=2𝑞²+2𝑞𝑡−𝑡²−3𝑞𝑘+𝑘²
𝑠=−𝑞²+2𝑞𝑡+2𝑡²−3𝑡𝑘+𝑘²
𝑦=𝑞²+𝑞𝑡+𝑡²−𝑘²
𝑥=𝑞²+𝑞𝑡+𝑡²−3(𝑞+𝑡)𝑘+2𝑘²

#Diophantine #equation #number-theory #holistic_algebra #diophantine

𝐴³+𝐵³=𝐶³+𝑄³

𝐴=(3𝑘−𝑡)((81𝑘⁵−108𝑘⁴𝑡+63𝑘³𝑡²−27𝑘²𝑡³+6𝑘𝑡⁴−𝑡⁵)𝑥²+3(9𝑘³−9𝑘²𝑡+3𝑘𝑡²−𝑡³)𝑥𝑦+(3𝑘−2𝑡)𝑦²)

𝐵=𝑡(3𝑘(27𝑘⁴−36𝑘³𝑡+21𝑘²𝑡²−6𝑘𝑡³+𝑡⁴)𝑥²+𝑡³𝑥𝑦−(3𝑘−2𝑡)𝑦²)

𝐶=(3𝑘−𝑡)(3𝑘(27𝑘⁴−36𝑘³𝑡+21𝑘²𝑡²−6𝑘𝑡³+𝑡⁴)𝑥²+(3𝑘−𝑡)³𝑥𝑦+(3𝑘−2𝑡)𝑦²)

𝑄=𝑡((−162𝑘⁵+216𝑘⁴𝑡−126𝑘³𝑡²+45𝑘²𝑡³−9𝑘𝑡⁴+𝑡⁵)𝑥²−3(18𝑘³−18𝑘²𝑡+6𝑘𝑡²−𝑡³)𝑥𝑦−(3𝑘−2𝑡)𝑦²)

#Diophantine #equation #number-theory #holistic_algebra #diophantine

𝐴³+𝐵³=𝐶³+𝑄³

𝐴=𝑘(3(𝑘³−2𝑡³)(𝑘²+𝑘𝑡+𝑡²)𝑥²+3(𝑘+𝑡)(𝑘³−2𝑡³)𝑥𝑦+(𝑘+𝑡)(𝑘²−𝑡²)𝑦²)

𝐵=𝑡(𝑘+𝑡)(3(𝑘⁴+𝑘²𝑡²+𝑡⁴)𝑥²+3𝑡³𝑥𝑦+(𝑡²−𝑘²)𝑦²)

𝐶=𝑘(𝑘+𝑡)(3(𝑘⁴+𝑘²𝑡²+𝑡⁴)𝑥²+3𝑘³𝑥𝑦+(𝑘²−𝑡²)𝑦²)

𝑄=𝑡(−3(2𝑘³−𝑡³)(𝑘²+𝑘𝑡+𝑡²)𝑥²−3(𝑘+𝑡)(2𝑘³−𝑡³)𝑥𝑦+(𝑘+𝑡)(𝑡²−𝑘²)𝑦²)

#Diophantine #equation #number-theory #holistic_algebra #diophantine

(𝐴−𝐵)³+𝐴³+(𝐴+𝐵)³=(𝐶−𝑄)³+𝐶³+(𝐶+𝑄)³

𝐴=64𝑘⁴𝑡³𝑥²+2𝑡𝑦²

𝐵=2𝑘(27𝑘⁶−18𝑘⁴𝑡²+20𝑘²𝑡⁴+8𝑡⁶)𝑥²+(9𝑘²+2𝑡²)(3𝑘²−2𝑡²)𝑥𝑦+2𝑘𝑦²

𝐶=8𝑡𝑘²(9𝑘⁴−4𝑘²𝑡²+4𝑡⁴)𝑥²+8𝑘𝑡(3𝑘²−2𝑡²)𝑦𝑥+2𝑡𝑦²

𝑄=64𝑘³𝑡⁴𝑥²+(3𝑘²−2𝑡²)²𝑥𝑦+3𝑘𝑦²

#Diophantine #equation #number-theory #holistic_algebra #diophantine

(𝐴−𝐵)³+𝐴³+(𝐴+𝐵)³=(𝐶−𝑄)³+𝐶³+(𝐶+𝑄)³

𝐴=2(36𝑎²−4𝑎𝑏+𝑏²)

𝐵=64𝑎²−𝑎𝑏+3𝑏²

𝐶=2(32𝑎²+𝑏²)

𝑄=74𝑎²−11𝑎𝑏+3𝑏²

A−B>0 and C−Q>0

𝐴=528𝑎²−40𝑎𝑏+𝑏²

𝐵=64𝑎²−25𝑎𝑏+3𝑏²

𝐶=128𝑎²+𝑏²

𝑄=764𝑎²−95𝑎𝑏+3𝑏²

#Diophantine #equation #number-theory #holistic_algebra

(𝐴−𝐵)³+𝐴³+(𝐴+𝐵)²=(𝐶−𝑄)³+𝐶³+(𝐶+𝑄)³

𝐴=4(24𝑎²−8𝑎𝑏+𝑏²)

𝐵=177𝑎²−59𝑎𝑏+6𝑏²

𝐶=4(33𝑎²−10𝑎𝑏+𝑏²)

𝑄=132𝑎²−49𝑎𝑏+6𝑏²

#Diophantine #equation #number-theory #holistic_algebra

𝐴³+𝐵³=𝐶³+𝑄³

𝐴=1144𝑎²−2024𝑎𝑏−11176𝑎𝑐+900𝑏²+9896𝑏𝑐+27300𝑐²

𝐵=−559𝑎²+989𝑎𝑏+5461𝑎𝑐−435𝑏²−4826𝑏𝑐−13335𝑐²

𝐶=273𝑎²−547𝑎𝑏−2731𝑎𝑐+269𝑏²+2726𝑏𝑐+6825𝑐²

𝑄=1092𝑎²−1928𝑎𝑏−10664𝑎𝑐+856𝑏²+9424𝑏𝑐+26040𝑐²

#Diophantine #equation #number-theory #holistic_algebra

The result of an attempt to solve this problem. There was the appearance of a new solution to a rather old problem.

https://math.stackexchange.com/questions/4591666/solving-cubic-systems-of-diophantine-equations

The problem of 4 cubes

𝐴³+𝐵³=𝐶³+𝑄³

𝐴=28𝑎²+12𝑎𝑏−68𝑎𝑐+2𝑏²−16𝑏𝑐+42𝑐²

𝐵=21𝑎²+𝑎𝑏−43𝑎𝑐−𝑏²+𝑏𝑐+21𝑐²

𝐶=42𝑎²+16𝑎𝑏−100𝑎𝑐+2𝑏²−20𝑏𝑐+60𝑐²

𝑄=−35𝑎²−15𝑎𝑏+85𝑎𝑐−𝑏²+17𝑏𝑐−51𝑐²

#Diophantine #equation #number-theory #holistic_algebra