A discussion about exponential diophantine equations and pythagorean triples

Question

Do all pythagorean triples $(a,b,c)$ have the identity that the three (exponential diophantine) equations

\begin{equation} a^x+b^y=c^z \end{equation} \begin{equation} b^y+c^z=a^x \end{equation} \begin{equation} a^x+c^z=b^y \end{equation} all have no more than one positive integer solution $(x,y,z)$?

This question is inspired by three exponential diophantine equations (1)$3^x+4^y=5^z$,(2) $4^y+5^z=3^x$, and (3) $3^x+5^z=4^y$.

Equations (1)(2)(3) all have no more than $1$ unique integer solution. Here, I'll give a proof for equation (3).

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Find all positive integer solutions $(x,y,z)$ of the equation $3^x+5^z=4^y$.

$\because 3^x+5^z=4^y$, $\therefore (-1)^x+1^z\equiv 0\;\; (\operatorname{mod} 4)$. Also, since $x,y,z$ are all positive integers, so $x$ must be a multiple of $2$. Let $x=2a$. Then $3^{2a}+5^z=2^{2y}$, so we can get $$5^z=\left(2^y+3^a\right)\left(2^y-3^a\right).$$ Suppose $2^y+3^a=5^m, 2^y-3^a=5^n\;\;(m,n\in\mathbb{N}, m>n)$, subtracting the two equations, we get $2\cdot3^a=5^m-5^n$. If $n\geqslant1$, then $5\mid 5^m-5^n$, so $5\mid 2\cdot3^a$, which is a contradiction. Therefore, $n=0$.

We have $2^y=3^a+1$. For $a=1$, we get $y=2$. For $a\geqslant 2$, writing the equation in modulo $4$, we get $0\equiv (-1)^a+1$, so $a$ must be odd. Let $a=2k+1$, then $3^{2k+1}+1=2^a$. i.e. $3\cdot 9^k+1=2^y$. Writing the equation in modulo $8$, we get $3\cdot 1^k+1\equiv 0$. Which is another contradiction, so $(a,y)=(1,2)$, which leads us to $x=2a=2$.

Plugging in $x=2, y=2$ into the original equation (equation (3)), we get $5^z=7$. Therefore, $z$ is not an integer, so the exponential diophantine equation $3^x+5^z=4^y$ doesn't have any solutions.

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I'm curious that is the above statement (at the beginning of the question) true for all pythagorean triples. e.g. Does the exponential diophantine equation $20^x+29^z=21^y$ have no more than one unique solution? Is there a general way to tackle with this problem?

Note:

1. Again, I hope my question is not a duplication.
2. $\mathbb{N}$ is the set of positive integers.

Answer 1

It's an interesting question.

For the particular equation $$ 3^x+5^z=4^y \qquad\qquad\;\;\;\, $$ your attempt to prove that there are no positive integer solutions had flaws, as was noted in the comments.

However for that equation, there are, in fact, no positive integer solutions.

This can be shown as follows . . .

Let $\mathbb{N}$ denote the set of positive integers and let $R,S,T$ be defined as \begin{align*} R&=\{3^x\;\text{mod}\;20{\,\mid\,}x\in\mathbb{N}\} \\[4pt] S&=\{5^z\;\text{mod}\;20{\,\mid\,}z\in\mathbb{N}\} \\[4pt] T&=\{4^y\;\text{mod}\;20{\,\mid\,}y\in\mathbb{N}\} \\[4pt] \end{align*} Then we get \begin{align*} R&=\{1,3,7,9\} \qquad\qquad\;\, \\[4pt] S&=\{5\} \\[4pt] T&=\{4,16\} \\[4pt] \end{align*} and it's then easily verified that there do not exist $r,s,t$ with $r\in R,\;s\in S,\;t\in T$ such that $$ r+s\equiv t\;(\text{mod}\;20) \qquad $$ Thus the congruence $$ 3^x+5^z\equiv 4^y\;(\text{mod}\;20) \;\;\, $$ has no positive integer solutions, hence the equation $$ 3^x+5^z=4^y \qquad\qquad\;\;\; $$ has no positive integer solutions.

Answer 2

An argument mod $3$ shows that $z$ is even then let $z=2{z_1}.$

Hence $3^x = 2^{2y} - 5^{2z_1} = ( 2^y + 5^{z_1})(2^y - 5^{z_1})$

$\bullet\ 2^y - 5^{z_1}\ne 1$

$2^y + 5^{z_1} \pmod 3 \implies z_1 \not\equiv y \pmod 3$
$2^y - 5^{z_1} \pmod 3 \implies z_1 \equiv y \pmod 3$
This is a contradiction.

$\bullet\ 2^y - 5^{z_1} = 1$

According to Catalan's conjecture, $2^y - 5^{z_1} = 1$ has no positive integer solutions.

Hence $3^x+5^z=4^y$ has no positive integer solutions.