Are there $a,b>1$ with $a^4\equiv 1 \pmod{b^2}$ and $b^4\equiv1 \pmod{a^2}$? 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.
 A: Maybe like this:
First observe that b^2 | (a^4-1) and a^2 | (b^4-1). Hence, we can write a^4-1=kb^2 and b^4-1=ma^2, with k, m being positive integers. 
So that gives then that a^4-1 = ((b^4-1)/m)^2-1=(b^8-2b^4+1)/m^2-1 = kb^2
That means that b^8-2b^4+1 = kb^2*m^2+m^2
Hence b^2(b^6-2b^2-km^2)=m^2-1
Now note that b^2 then should be a divider of m^2-1, cause b^6-2b^2-km^2 is still a integer.
That means that m^2 = 1 mod b^2, with m and b being still integers.
Hence, m^2 = 1 +lb^2, with l being integer. 
Substitution then gives that b^6-(2+lk)b^2-(k+l)=0
Now solve this equation for b^2 and observe that this gives two complex solutions and one real solution, the only thing to do is to solve that explicitly and try for integer values of l and k to get integer value of b, thereby m^2 also becomes integer and a divider.
A: 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!
