Are there $a,b>1$ with $a^4\equiv 1 \pmod{b^2}$ and $b^4\equiv1 \pmod{a^2}$?

Question

Are there solutions in integers $a,b>1$ to the following simultaneous congruences? $$ a^4\equiv 1 \pmod{b^2} \quad \mathrm{and} \quad b^4\equiv1 \pmod{a^2} $$

A brute-force search didn't turn up any small ones, but I also don't see how to rule them out.

Answer 1

I would think this could be attacked as follows. Write $b = b_1b_2b_3$ for integers $b_1,b_2,b_3$, such that $b_1^2 \mid (a-1)$ and $b_2^2 \mid (a+1)$ and $b_3^2 \mid (a^2+1)$. Hence the first condition is satisfied, i.e.,

$$ a^4-1 = (a-1)(a+1)(a^2+1) \equiv 0 \pmod{b^2}.$$

Write $a-1 = c_1b_1^2$ and $a+1=c_2b_2^2$ and $a^2+1=c_3b_3^2$ for integers $c_1, c_2,c_3$.

From this point, I would think there would be a number of interesting ways to solve or contradict the second condition

$$ b^4-1 = (b-1)(b+1)(b^2+1) \equiv 0 \pmod{a^2}.$$

Good luck!