由a-b=2, 得知a=b+2
a2+b2=2b2+4b+4
由除以8有餘數, 得知f(b)=8P+R
(2b2+4b+4)=8P+R=2[(4P)+(1/2)R]
=2(b2+2b+2)
=2[(b+1)2+1], b+1為(質數+1)=偶數, 偶數平方必為4的倍數
=2(4P+1)
(1/2)R=1, R=2