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

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.

• Dec 26, 2019 at 17:06
• This appears to be a G research December puzzle, and it looks hard: The $a=2$ case is an open problem from at least year 2014 (or earlier). Dec 27, 2019 at 11:19

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 $$b = 386$$.

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 $$k$$).

• I claimed that I was unable to find any more solutions, as opposed to there being no more. But thank you for showing that particular solution, I had not found that.
– J-S
Dec 26, 2019 at 17:09
• FYI: even for $a = 2$ there are more: $b = 20136, 59140, 373164544$. See oeis.org/… Dec 26, 2019 at 17:11
• Also, for $a = 3$ there are $b = 27, 19683$. This is not yet on OEIS. Dec 26, 2019 at 17:14

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 $$x^p+1$$.

• I don't know why, but I already gave this example (in a more general form) in my answer. Dec 26, 2019 at 17:37