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

@geekbrit pulling a #Diophantine on unsuspecting public, I see 🤣

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

A funny discussion of the solution of one Diophantine equation was singled out separately.

New ideas are very hard to make their way. Although the ideas of holistic algebra are not new. They were just hammered and closed.

Although you can set it simply. How to write a formula for such a Diophantine equation?

\( ax^2+by^2+cz^2=dv^2+tq^2 \)

#Diophantine #equation #number-theory

https://chat.stackexchange.com/rooms/141036/discussion-on-answer-by-individ-diophantine-equation-x-2y-2z-2m

#introduction

I'm an assistant professor at @WMI_research. I study equations and inequalities in whole numbers, using #lattices and #fourier #analysis.

I’m also dabbling in something very different: nonlinear dispersive partial differential equations, a type of mathematical model for real things from ocean waves to fibre optic cables.

#maths #math #research #teaching #HigherEducation #WelcomeToMathstodon #Academia #Science #universities #mathed #NumberTheory #diophantine #mathematics #numbers

@jbeardsleymath I think Pythagorean triples are just super cool. When I checked wikipedia to make sure that a² + b² = c² is a #Diophantine equation (it is), there was a potentially-interesting list of others?

About Pythagorean triples, I was astounded to find out there's a *formula* for them. It's not even that complicated! One method of proof is to find rational points on the circle, and it boils down to the quadratic formula (really). I found that to be amazing!

Last couple of nights I've been falling asleep on another little mental calculation.

Oh yeah, it's time for another episode of #CountingSheep, bay-bee!

This time we're doing linear #Diophantine equations, i.e. the Euclidean algorithm.