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发帖数: 204 | 1 If a^2=b(mod p) and c^2=b+1(mod p), then c^2-a^2=1(mod p).
Let d=c+a, e=c-a, then d*e=1(mod p). given any d!=0(mod p),
there have a unique e!=0(mod p) such that d*e=1(mod p).
If d-e!=0(mod 2), then choose a appropriate e(such as e+p) to
make d-e=0(mod 2). then you can solve a,c(mod p).
The counting of solotions of a,c(mod p) then becomes almost trivial.
Notice both a^2=(p-a)^2(mod p), so you need consider how many times
you over-counted b,b+1(mod p), then devided it, to get the number
of solutio |
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