if p is prime and a^2≡b^2(modp)
show that either p|(a+b) or p|(a-b)
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ith equality iff
{r − a1, r − a2, . . . , r − a2n} = {1, 2, . . . , n, −1, −2, . . . , −n}.
Therefore, this must be the case, so
(r − a1) + (r − a2) + · · · + (r − a2n)
= 2nr − (a1 + a2 + · · · + a2n)
= 1 + 2 + · · · + n + (−1) + (−2) + · · · + (−n) = 0
⇒ r =
a1 + a2 + · · · + a2n
2n
.
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