How to solve these diophantine equations?

Question

Show that the diophantine equations $(a^{p^2} + b^{p^2}) - (a^p + b^p)\equiv 0 \pmod {p^4}$ and $(a^p + b^p)^{p-1}\equiv 1 \pmod {p^2}$ has no solution for $(a,p) = (b,p) = (a,b) = 1$, $p$ is an odd prime, $a$ is even, $b$ is odd and both $a,b >= 2$

Answer 1

Since those are all congruences, conditions such as $(a,b)=1$ or $a,b\geq 2$ don't really make sense (that is, they don't help in avoiding the trivialities you want to exclude, such as those coming from $a=1, b=1$). For instance, write $a=p^4m+1$ and $b=p^4n+1$. Any such choice produces a solution but it happens very often that those two are coprime. For example $a=2p^4+1$ and $b=p^4+1$ are solutions satisfying all your conditions.