By the problem:
Since s^2 leaves a remainder of 1 upon division by 4, it follows that s must be odd.
Let s = 2q+1. Then, we have:
q^2 + q = p^4+ p^3+ p^2+ p...........(#)

Now, (p^2 + p/2 – 1/2)(p^2+p/2 + 1/2)
= p^4 + p^3 + 1/4*(p^2)- 1/4
< p^4 + p^3 + p^2+ p, whenever:
3/4*p^2 + p + 1/4 > 0
Or, 3*p^2 + 4p + 1> 0
Or, (3p+1)(p+1) > 0
This is satisfied for all values of p with the exception of -1< = p< = -1/3

Also, (p^2 + 1/2*p)(p^2 + 1/2*p + 1)
= p^4+p^3+5/4*p^2 + 1/2*p, which is greater than:
p^4 + p^3 + p^2 + p, whenever:
1/4*p^2 - 1/2*p> 0
Or, p(p-2) > 0
This is satisfied for all values of p with the exception of 0< =p< =2

If possible, let there exist integers which are greater than A = p^2+p/2 - 1/2 but less than p^2 + p/2 = B.
For even p, B is an integer and A = B -1/2
For odd p, A is an integer and B = A+1/2
Hence, there does not exist integers which are greater than (p^2 + p/2 - 1/2) and less than (p^2 +p/2)

Similarly, it can be shown that there does not exist any integer which is greater than
(p^2 + p/2 + 1/2) and less than (p^2 + p/2 + 1)

Accordingly, integer solutions to (#) is only possible whenever:
(-1<=p<=-1/3) U (0<=p<=2); or, -1<=p<=2
From (#), we observe that
p=-1 gives q^2+q = 0, so that q= 0, -1
p = 0 yields q = 0, -1
p = 1 yields q^2 + q- 4 = 0, which do not possess integer solutions in q.
p = 2 gives q^2 + q - 30 = 0, so that (q+6)(q-5) = 0 giving q = -6, 5
Therefore , (p, q) = (-1, 0), (-1, -1), (0, 0), (0, -1), (2, -6) and (2, 5) and, consequently:
(p, s) = (-1, 1); (-1, -1); (0, 1); (0, -1); (2, -11); (2, 11) constitutes all possible solutions to the given problem.

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