# $7^x-3^y=4$ 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

• I remember this as an Indian National Math Olympiad problem – N.S.JOHN Oct 20 '18 at 20:08
• I found out that $y$ must be of the form $6k+1$. Don't know if it will be of any help – Jakobian Oct 20 '18 at 20:24

## 2 Answers

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.

• Looks good to me, +1 – Jakobian Oct 20 '18 at 21:38
• I didn't quite get why $7^x\equiv7^5$ and why $3^y\in\lbrace7,11\rbrace\mod13$ – Bruno Andrades Oct 20 '18 at 23:27
• @BrunoAndrades I've added a bit more explanation to my post. – Carl Schildkraut Oct 20 '18 at 23:43

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 $$x=y=1$$.