A problem by Amanda Hm

Question: 1 + p + $p^{2}$ + $p^{3}$ + $p^{4}$ = $n^{2}$

p + $p^{2}$ + $p^{3}$ + $p^{4}$ = $n^{2}$ - 1 = (n+1)(n-1)

n = ±1 (mod p)

Objective: Bound as such $p^{2}$ < $n^{2}$ < $q^{2}$ so that we can find the values of n

( $p^{2}$ + 1)^2 < 1 + p + $p^{2}$ + $p^{3}$ + $p^{4}$ Expand and observe that it stands for all natural p (greater than or equal to one).

1 + p + $p^{2}$ + $p^{3}$ + $p^{4}$ < ( $p^{2}$ + p - 1)^2

( $p^{2}$ + p - 1)^2 is chosen because it is the next square that satisfies n = ±1 (mod p) Expand and observe that it stands for all natural p greater than 3.

Therefore, we know that p needs to be 3 or smaller. Since p is prime, just check if p = 2, p = 3 works. Then observe that (p,n) = (3,11) is the only solution.