Does $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ have any solutions where $a$ and $b$ are co-primes less than $p$? How will you prove that $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ has no solution where $p$ is a prime number and $a$, $b$ are two co-primes less than $p$? If this equation has a solution, then what it is it?
Edit- Some solutions considered taking $a$ or $b$ as 1. Are there any solutions where neither $a$ nor $b$ is 1? 

For $p=2$ it says $a^2 + b^2 \equiv 1 \mod 4$. This can only be achieved if one of $a$ and $b$ is divisible by 2, so this one doesn't have any solutions.
For $p=3$, one has $(a^3+b^3)^2 \equiv 1 \mod 9$, so $a^3+b^3 \equiv 1 \mod 9$ or $a^3+b^3 \equiv 8 \mod 9$. The former has no solutions, same argument as for $p=2$. But third powers are only congruent to 0,1,8 mod 9, so it won't work without using a 0. 
For $p=5$ one can verify in the same way that there are no solutions. 
For $p=7$ one has $a=1$, $b=2$ as a solution as well as 97 other pairs.
