$7^x-3^y=4$ in the integers 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
 A: 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. 
A: 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 $x=y=1$. 
