$2^{x}+ 5^{y}= 7^{z}$

Question

Help me solve this integers exponents equation:

$$2^{x}+ 5^{y}= 7^{z}$$

We have: $z= \frac{i(2\,\pi\,n-i\,\log(2^{x}+ 5^{y}))}{\log(7)}$

I ask because I read that a slightly complication of the equation occuring in Fermat's last theorem can lead to an undecidable case and I wonder whether the given form already is sufficient to achieve this.

Answer 1

Only $x=y=z=1$, corresponding to $2+5=7$, can work. Most of this proof rehashes the comments, the real purpose here is to consolidate all the steps into one place.

1) $x$ is odd, or else $2^x \equiv 7^z \equiv 1 \bmod 3$ and $5^y$ can't be divisible by $3$.

2) $y$ is odd, or else $5^y \equiv 7^z \equiv 1 \bmod 3$ and $2^x$ can't be divisible by $3$.

3) If $x$ and $y$ are both odd and $x\ge 3$, then $2^x+5^y \equiv 0+5=5 \bmod 8$ but $7^z \in \{1,7\} \bmod 8$. This forces $x=1$.

4) If $x=1$ and $y\ge 2$, then $2^x+5^y \equiv 2+0=2 \bmod 25$ but $7^z \in \{1,7,18,24\} \bmod 25$. This forces $y=1$ leaving $2+5=7$ as the only possible equality.

Answer 2

This is the answer attributed to @Barry Cipra and a little from myself.

Firstly, $(-1)^x+(-1)^y\equiv 1\pmod 3$, meaning that $x,y$ are both odd

Secondly, if $x>1$, then $1=7^z\pmod 4$, meaning that $z=2k$ for some $k$.

Thirdly, modulo $15$ gives $2^x+5\equiv 4^k\pmod{15}$, but $$2^1=2, 2^2=4^1=4, 2^3=8, 2^4=4^2=1\pmod{15}$$, so it’s a contradiction. Hence $x=1$ and $z$ is odd.

Forth, modulo $25$ gives, if $y>1$, $2\equiv 7^z\pmod{25}$, but this also cannot happen. And we are done

Answer 3

This is just a variant on the ideas in BAI's answer.

Working mod $3$ we have $2^x+2^y\equiv1$, from which it follows that $x$ and $y$ are both odd.

Working mod $5$ we now have $2^x\equiv2^z$ (since $y\not=0$), from which it follows that $x$ and $z$ have the same parity. Consequently $x$, $y$, and $z$ are all odd (as stated in BAI's original comment beneath the OP!).

Working mod $4$ and writing $z=2k+1$, we have $2^x+1\equiv3^{2k+1}\equiv3$, so we must have $x=1$.

Working mod $25$, we now have $2+5^y\equiv7^{2k+1}\equiv(-1)^k7$, from which it follows that $y=1$ (and $k$ is even, not that that matters much), since $2\not\equiv\pm7$ mod $25$.

It now follows that $2^1+5^1=7^1$ is the only solution.