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I have to find all solutions to $7^x-3^y=4$ in the integers, I've already proven that $x$ and $y$ have the same parity and that they cannot be even. But I'm stuck in the case when $x$ and $y$ are odd. Could someone show me how to solve for all solutions? I know there is at least one, $(x,y)=(1,1)$, but I don't know if there's any other solution

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  • $\begingroup$ I remember this as an Indian National Math Olympiad problem $\endgroup$ – N.S.JOHN Oct 20 '18 at 20:08
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    $\begingroup$ I found out that $y$ must be of the form $6k+1$. Don't know if it will be of any help $\endgroup$ – Jakobian Oct 20 '18 at 20:24
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If $y\leq 1$, the only solution is $(1,1)$, which you found. If $y\geq 2$, then consider the equation $\bmod 9$. We see

$$7^x\equiv 4\bmod 9$$

We have

$$7^0\equiv 1,\ 7^1\equiv 7,\ 7^2\equiv 4,\ 7^3\equiv 1,$$

so $7^x\equiv 4\bmod 9$ iff $x\equiv 2\bmod 3$. As a result, we have $x\equiv 5\bmod 6$ (as you have already shown $x$ is odd). So, as

$$7^{12}\equiv 1\bmod 13,$$

we have

$$7^x\equiv 7^5\mathrm{\ or\ }7^{11}\equiv \pm 2\bmod 13$$

since $x\equiv 5\bmod 6\implies x\equiv 5\mathrm{\ or\ }11\bmod 12$. This means that

$$3^y=7^x-4\equiv (\pm 2)-4\in\{7,11\}\bmod 13.$$

However

$$3^0\equiv 1,3^1\equiv 3,3^2\equiv 9,3^3\equiv 1\bmod 13,$$

so we get a contradiction.

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  • $\begingroup$ Looks good to me, +1 $\endgroup$ – Jakobian Oct 20 '18 at 21:38
  • $\begingroup$ I didn't quite get why $7^x\equiv7^5$ and why $3^y\in\lbrace7,11\rbrace\mod13$ $\endgroup$ – Bruno Andrades Oct 20 '18 at 23:27
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    $\begingroup$ @BrunoAndrades I've added a bit more explanation to my post. $\endgroup$ – Carl Schildkraut Oct 20 '18 at 23:43
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We first show that if x and y have a common divisor like c, then a number such as $a^x-b^y$ has a factor like $a-b$(it does not mean if x and y have no common divisor $a^x-b^y$ definitely has no common divisor like $a-b$):

We know that:

$(a^x-1, a^y-1)=a^c-1$

$(b^x-1, b^y-1)=b^c-1$

Now take one say $a^x-1$ which has a factor like $a^c-1$ and $b^y-1$ which has a factor like $b^c-1$ and subtract them:

$gcd(a^x-b^y)≡a^c-b^c=(a-b)(a^{c-1}+a^{c-2}b+ . . .)$

So we may write:

$a^x-b^y=k(a-b)$

It is clear that for a certain set of values for a, b,x and y there exist only one k to satisfy this relation. Hence there are infinitely many such unique relation for certain values of involved parameters. That means equation $7^x-3^y=4$ in which $a=7,. b=3,. $ and $k=1$ are certain values has only one solution $x=y=1$.

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