dp?12??1(modp).
3. 若素数p?2,证明:?1是模p的二次剩余的充要条件是p?1(mod4).
??p?1??当p?1(mod4)时,????!???1(modp).
2????证明:由Euler判别法知道?1是模p的二次剩余的充要条件是(?1)又p?2,所以(?1)p?12p?122?1(modp).
?1.即p?1(mod4).
由Wilson定理,(p?1)!??1.
p?1p?1?1?(p?1)!?(?)(??1)222p?1p?1p?1??p?1??(?1)1(?1)?(?1)2???!?(modp)
22??2??2??p?1??当p?1(mod4)时,????!???1(modp).
??2??4.设p是奇素数,证明:1,2,?j2?p(p?1),p?1中全体模p的二次剩余之和S??p???.
24j?1?p?p?122p?12?j2?p(p2?1)p(p?1)由此可以证明当p?1(mod4)时,p?????.
244j?1?p?证明:因为d是模p的二次剩余当且仅当d?1,2,(p?1p?12?1)2,()(modp). 22?j2?p?1设j?pqj?rj(1?rj?p),1?j?.则qj???.
2?p?2p?12j?1p?12j?1于是S??r??jp?1p?1p?1p?1(?1)(p?1?1)22222?2??j??j?p(p?1)j222j?p?????p?????p???.
624j?1?p?j?1?p?j?1?p?p?12?j2?p(p2?1)?S. 若p?1(mod4),则p????24j?1?p?因为p?1(mod4),由Euler判别法知道rj与p?rj同为二次剩余或非二次剩余. 又在模p的一个既约剩余系中,恰好有
p?12pp?1p(p?1)p?1个模p的二次剩余,所以S???.
2242?j2?p(p2?1)p(p?1)于是p?????.
p244j?1??p?12