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.
素数平方余数
记pn是第n个素数：2、3、5、7、11……；记r是(pn?1)n + (pn+1)n被pn2除所得的余数。
例如，当n = 3时，p3 = 5，而43 + 63 = 280 ≡ 5 mod 25。
使余数首次超过109的最小n值是7037。
求使余数首次超过1010的最小n值。