Find all integers $a$ and $b$, such that $\frac{a^b + 1}{b^a + 1}$ an integer.

Question

I've recently been attempting to solve the following problem and made some initial progress. However, im not able to progress any further, any hint would be greatly appreciated!

Let $a$ and $b$ be positive integers. For which values of $a$ and $b$ is the quotient $\frac{a^b + 1}{b^a + 1}$ an integer?

Thus far I have found that when $b=1$, then $\frac{a^b + 1}{b^a + 1}$ is an integer for all odd positive integers $a$. Likewise, the quotient is an integer when $a=b$ and when $(a,b)=(2,4)$ and $(a,b)=(4,2)$.

I haven't yet been able to find any more solutions and I also have no ideas on how I can prove that there are no more solutions, other than the ones given.

Answer 1

I'm not sure where you got this problem. It seems to me that it is either impossible or very difficult to get a complete answer.

In particular, your claim that "there are no more solutions" is wrong.

As an example: $a = 2$ and $b = 386$.

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There are other families of examples: take any odd $a$ and take $b = a^n$, where $n$ divides $a^{n - 1}$ (e.g. $n = a^k$ for some $k$).

Answer 2

There are infinitely many solutions of the following form.

If $p$ is an odd positive integer and $k = p^{p^r}$, then $k^p+1 = p^{p^{r+1}}+1$ divides $p^k+1 = p^{p^{p^r}}+1$. This is because $x+1$ divides $x^p+1$.