3^x + 4^y equals 5^z has Unique Solution

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Theorem

The Diophantine equation:

$3^x + 4^y = 5^z$

has exactly one solution in (strictly) positive integers:

$3^2 + 4^2 = 5^2$


Proof

Rewriting our Diophantine equation modulo $4$ we have:

$\paren {-1}^x + 0^y \equiv 1^z \pmod 4$

Therefore $x$ is even.

Rewriting our Diophantine equation modulo 3 we have:

$0^x + 1^y \equiv \paren {-1}^z \pmod 3$

Therefore $z$ is even.


Since $x$ and $z$ must both be even, we will rewrite $x$ as $2 r$, $z$ as $2 s$ and notice that $4$ is $2^2$.

We now have:

\(\ds 3^{2 r} + 2^{2 y}\) \(=\) \(\ds 5^{2 s}\)
\(\ds 5^{2 s} - 3^{2 r}\) \(=\) \(\ds 2^{2 y}\) rewriting the equation
\(\ds \paren {5^s + 3^r} \paren {5^s - 3^r}\) \(=\) \(\ds 2^{2 y}\) factoring the left hand side: Difference of Two Squares


We can see that the right hand side is a product consisting entirely of instances of $2$.

Therefore:

$\paren {5^s + 3^r} = 2^u$

and:

$\paren {5^s - 3^r} = 2^v$

where $u > v$ and $u + v = 2 y$.


Solving for $5^s$, we add the two equations and get:

\(\ds \paren {5^s + 3^r} + \paren {5^s - 3^r}\) \(=\) \(\ds 2^u + 2^v\)
\(\ds 5^s\) \(=\) \(\ds 2^{u - 1} + 2^{v - 1}\) Dividing both sides by $2$
\(\ds \) \(=\) \(\ds 2^{v - 1} \paren {2^{u - v} + 1}\) Factoring out $2^{v - 1}$


Solving for $3^r$, we subtract the two equations and get:

\(\ds \paren {5^s + 3^r} - \paren {5^s - 3^r}\) \(=\) \(\ds 2^u - 2^v\)
\(\ds 3^r\) \(=\) \(\ds 2^{u - 1} - 2^{v - 1}\) dividing both sides by $2$
\(\ds \) \(=\) \(\ds 2^{v - 1} \paren {2^{u - v} - 1}\) factoring out $2^{v - 1}$


Since both $5^s$ and $3^r$ are odd, $v$ must be equal to $1$.

Let $t = u - v$.

We now have:

$5^s = \paren {2^t + 1}$
$3^r = \paren {2^t - 1}$


Let us now consider the second equation Modulo 3:

$0 \equiv \paren{-1}^t - 1 \pmod 3$

Therefore t must be even.

We will rewrite the second equation above with $t$ as $2 w$:

\(\ds 3^r\) \(=\) \(\ds \paren {2^{2w} - 1}\)
\(\ds \) \(=\) \(\ds \paren {2^w + 1} \paren {2^w - 1}\) Difference of Two Squares


We can see that the left hand side is a product consisting entirely of instances of $3$.

Therefore:

$ 3^m = \paren {2^w + 1}$

and:

$ 3^n = \paren {2^w - 1}$

where $m > n$ and $m + n = r$.


Subtracting the $2$ equations above we get:

\(\ds 3^m - 3^n\) \(=\) \(\ds \paren{2^w + 1} - \paren{2^w - 1}\)
\(\ds \) \(=\) \(\ds 2\)

Therefore $m = 1$ and $n = 0$.


From above:

$3^m = \paren {2^w + 1}$

therefore:

$w = 1$


Recall that $t = 2w$.

Therefore:

$t = 2$

and:

$5^s = \paren {2^t + 1}$
$3^r = \paren {2^t - 1}$

Therefore $r = 1$ and $s = 1$.


Finally recall that $x = 2 r$ and $z = 2 s$.

Therefore we arrive at the only possible solution in (strictly) positive integers:

$x = 2$, $y = 2$ and $z = 2$

$\blacksquare$


Sources

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