Math, asked by manojshivajihase04, 7 months ago

if p is prime and a^2≡b^2(modp)
show that either p|(a+b) or p|(a-b)

Answers

Answered by padmamaloth1986
6

{\huge{\underline{\underline{\mathfrak{\pink{A}\green{n}\blue{s}\red{w}\orange{e}\pink{r}}}}}}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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