# Problem 123: Prime square remainders
Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder
when (pn−1)n + (pn+1)n is divided by pn2. For example, when n = 3, p3 =
5, and 43 + 63 = 280 ≡ 5 mod 25. The least value of n for which the
remainder first exceeds 109 is 7037. Find the least value of n for which
the remainder first exceeds 1010.