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Math 110 Problem Set 2 Solutions - Stanford University

Math 110 Problem Set 2 3 be prime. Show that the only Solutions tox2 1 (modp) arex 1 (modp).Solution:Note that we have(x+ 1) (x 1) x2 1 0 (modp)Thus, by Exercise 7(a), we have that eitherx 1 0 (modp) orx+ 1 0(modp) so we havex 1 (modp) as 2 (mod 7) andx 3 (mod 10). What isxcongruent to mod70?Solution:This is easiest to do by trial and error. The 2nd condition meansthatxis 3, 13, 23, 33, 43, 53, or 63 modulo 70. Checking, we see that the onlyone of these which is 2 modulo 7 is 23, soxmust be 23 (mod 70). Note thatthe Chinese remainder Theorem guarantees that there is a unique answer tothe prime.

Math 110 Problem Set 2 Solutions ... the Chinese Remainder Theorem guarantees that there is a unique answer to the question. 3.11 Let p be prime. Show that ap · a (mod p) for all a. Solution: Let us consider two cases: p divides a, and p does not divide a. If p divides a, then both sides are 0 modulo p.

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