Problem 273
Sum of Squares
Consider equations of the form: a2 + b2 = N, 0 ≤ a ≤ b, a, b and N integer.
For N=65 there are two solutions:
a=1, b=8 and a=4, b=7.
We call S(N) the sum of the values of a of all solutions of a2 + b2 = N, 0 ≤ a ≤ b, a, b and N integer.
Thus S(65) = 1 + 4 = 5.
Find ∑S(N), for all squarefree N only divisible by primes of the form 4k+1 with 4k+1 < 150.
平方的和
考虑如下形式的方程：a2 + b2 = N, 0 ≤ a ≤ b, a、b和N均为整数。
对于N=65，方程有两个解：
a=1, b=8 和 a=4, b=7。
对于方程a2 + b2 = N, 0 ≤ a ≤ b, a、b和N均为整数，所有的解中的a值的和记为S(N)。
因此S(65) = 1 + 4 = 5。
有一些无平方因子数N，只能被形如4k+1且< 150的素数整除，对所有这样的数求∑S(N)。