<s>[INST] Existence of Solutions to a $2-$Equation System of Congruences Do there exist $a, b> 1$, such that
$$ a^4 \equiv 1 \pmod{b^2}$$ and $$ b^4 \equiv 1 \pmod{a^2}.$$
 [/INST] If the given $2$ relations hold true, then 
$$b^2\big|a^4-1 \Rightarrow b^2\big|(a^2+1)(a+1)(a-1)$$
and 
$$a^2\big|b^4-1 \Rightarrow a^2\big|(b^2+1)(b+1)(b-1)$$
Also we have the relations below for suitable positive integers $k,l$ that
$$a^4=kb^2+1$$
$$b^4=la^2+1$$
Hence $$a^4-b^4=kb^2-la^2$$ or $$a^2(a^2+l)=b^2(b^2+k)$$
For this to hold, $a^2$ divides both L.H.S. and R.H.S.
But $a^2$ and $b^2$ do not divide each other, as per the given relations.
So $$a^2\big|b^2+k$$ and similarly $$b^2\big|a^2+l$$
Since $k,l$ are positive, $$a^2\big|b^2+k \Rightarrow b^2>a^2$$ and similarly $$b^2\big|a^2+l \Rightarrow a^2>b^2$$
Both of these cannot hold true together.
Hence such $a,b$ do not exist.
</s>