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I don't know how to solve $7^x-3^y=4$... I tried to see something $\pmod 7$ and $\pmod 3$ but it doesn't help at all. Can anyone give me some hints about it?

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    $\begingroup$ $x$ and $y$ are supposed to be integers, aren't they? $\endgroup$
    – ajotatxe
    Commented May 9, 2019 at 18:08
  • $\begingroup$ yes they are... I found (1,1), but then? How to find other solutions? $\endgroup$ Commented May 9, 2019 at 18:10
  • $\begingroup$ See here, with $z=2$. This is a similar case. Actually, I found a solution here, in "Art of Problem Solving". $\endgroup$ Commented May 9, 2019 at 18:11
  • $\begingroup$ According to this the only positive integers $n$ such that $n$ and $n+4$ are both perfect powers are $4,32$ and $121$ none of which are of the form you want. I'm not sure, though, whether the results in that table reflect proven theorems or just the current state of numerical searches. $\endgroup$
    – lulu
    Commented May 9, 2019 at 18:12
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    $\begingroup$ @lulu there is an elementary method when we have primes $p,q$ and $p^m - q^n = c.$ math.stackexchange.com/questions/1941354/… and math.stackexchange.com/questions/1946621/… $\endgroup$
    – Will Jagy
    Commented May 9, 2019 at 18:20

1 Answer 1

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Note: these are not quite the same $x,y$ as in the question.

brief version, we reach $$ 7(7^x-1) = 3(3^y - 1) \; , $$ assume that $x,y > 0$ and produce a contradiction.

As $7 | (3^y - 1)$ so that $$ 3^y \equiv 1 \pmod 7 \; , $$ we find $$ 6 | y $$

Then $(3^6 - 1 )| (3^y - 1)$ while $$ 3^6 - 1 = 8 \cdot 7 \cdot 13 $$

Next $13 | (7^x - 1)$ so that $$ 7^x \equiv 1 \pmod {13}, $$ so $12|x $ and we use $3 | x.$ Then $7^3 - 1 | 7^x - 1,$ while $$ 7^3 - 1 = 2 \cdot 9 \cdot 13. $$

We have reached $$ 9 | 3 (3^y-1) \; , \; $$ or $$ 3 | (3^y-1) \; , \; $$ which contradicts $y > 0.$

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