2016/120) Solve in positive integers $p^x = y^4 + 4$ and p is prime

factor the RHS ro get $p^x = (y^2-2y+2)(y^2+2y+2)$
both are power of p and $y^2-2y+2)< y^2 + 2y + 2$ so $y^2-2y + 2$ is a factor of difference that is 4y
so $y^2-2y+2$ is a factor of $4y^2$ or $4y^2-4(y^2-2y+2) = 8(y-1)$
so $y^2-2y+2$ is a factor of 8 that is it is 1 or 2 or 4 or 8
if it is 1 then y = 1, p = 5 and x = 1
if it is 2 then y=2 giving $p^x = 20$ which is not possible and if it is 4 or 8 we do not have integer solution
so only solution is $p = 5, x = y= 1$