If p satisfies the compound inequality (2p^2 + p + 1)^2 > 4p^4 + 4p^3 + 4p^2 + 4p + 1 > (2p^2 + p)^2 then it is obvious that the middle expression is not a square since it is strictly between two consecutive squares.

The only intervals where the compound inequality fails are [-1, -1/3] and [0, 2]. There are only four integers in this range: -1, 0, 1, and 2.

Direct calculation shows that p=1 is not a solution and p=-1 and p=0 are solutions with s = +/-1 and p=2 is a solution with s=+/-11.

The complete solution set is (-1, 1), (-1, -1), (0, 1), (0, -1), (2, 11), (2, -11).