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.

Solution 1:

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.