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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.

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    There is at least one solution given by $x=y=z=1$. What are you exactly looking for? – Gibbs May 12 '18 at 12:21
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    Well, $x,y,z$ must be all odd – BAI May 12 '18 at 12:26
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    @BAI, why is that? If $x$, $y$, and $z$ are all odd, then you can conclude $x=1$, since $5^{odd}\equiv5\not\equiv7\equiv7^{odd}$ mod $8$. Come to think of it, if $x\gt1$, then you need $z$ to be even, since $5^y\equiv1$ mod $4$. – Barry Cipra May 12 '18 at 12:29
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    @BarryCipra by modulo $3$ we get $(-1)^x\equiv 1-(-1)^y\pmod 3$, so $y$ is odd and so does $x$, and modulo $4$ we come to $2+1\equiv(-1)^z\pmod 4$ so $z$ is odd as well. – BAI May 12 '18 at 12:32
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    @BAI, ah, I see what you mean: $x$ and $y$ must always be odd. But you're wrong about $z$, since $7^z\equiv1^z\equiv1$ mod $3$ for all $z$. – Barry Cipra May 12 '18 at 12:36
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    @BarryCipra well, I came to $z$ in modulo $4$ in my comment – BAI May 12 '18 at 12:37
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    @BarryCipra and I think we’ve done it: from our previous ones we know $x=1$, and if $y>1$, then $7^z\equiv 2\pmod 25$, which cannot be happening. – BAI May 12 '18 at 12:39
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    @BAI, I'm not sure what you mean (in your "modulo $4$" comment). Working mod $4$, I showed $z$ must be even unless $x=1$. – Barry Cipra May 12 '18 at 12:39
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    @BAI, I agree, we're close to done. Go ahead and post it as an answer! – Barry Cipra May 12 '18 at 12:41
  • @BarryCipra sorry I think I suddenly lost my mind $\pmod 4$ in the previous comment... that is wrong. – BAI May 12 '18 at 12:44
  • @BAI, no sweat, I suffer lapses of logic all the time. It does occur to me that one case is going to remain: showing $2=7^z-5^y$ has only $y=z=1$ as a solution, which sounds Catalan-equationesque. – Barry Cipra May 12 '18 at 12:46
  • @BarryCipra well, I’ve answered this case ($7^z-5^y=2$) above by modulo $25$. Just get headache about fixing the mod 4 flaw – BAI May 12 '18 at 12:49
  • @BAI, ah!!! (See what I mean about my suffering lapses of logic?) Please post the answer! – Barry Cipra May 12 '18 at 12:53
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    According to the Beal Conjecture, this has no solution for x, y, z > 2 because 2, 5 and 7 do not share a common prime factor – Phil H May 12 '18 at 13:01
  • To get an undecidable statement, you have to add a parameter, something like $2^x+5^y+7^z=a$ where $a$ is arbitary , but fixed and the question is whether integers $x,y,z$ exist solving the equation. – Peter May 12 '18 at 13:02
  • @Peter, why can't the parameter occur as an exponent, something like $2^x+5^y=7^a$ where $a$ is arbitrary but fixed and the question is whether integers $x$ and $y$ exist solving the equation? – Barry Cipra May 12 '18 at 13:22
  • @BarryCipra This would work as well, but we need some parameter because otherwise, by definition the problem is always decidable. – Peter May 12 '18 at 13:25

3 Answers3

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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.

Oscar Lanzi
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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.

Barry Cipra
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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

Oscar Lanzi
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BAI
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