Question

Prove: If p is prime and p ≡ 7 (mod 8), then p |
2^{(p−1)/2} − 1. (Hint: Use the Legendre symbol (2/p) and
Euler's criterion.)

Answer #1

Let p be be prime and p ≡ 1 (mod 4|a|). Prove that a is a
quadratic residue mod p.

Let p be a prime that is congruent to 3 mod 4. Prove that there
is no solution to the congruence x2≡−1 modp. (Hint: what would be
the order of x?)

Let p be an odd prime, and let x = [(p−1)/2]!. Prove that x^2 ≡
(−1)^(p+1)/2 (mod p).
(You will need Wilson’s theorem, (p−1)! ≡−1 (mod p).) This gives
another proof that if p ≡ 1 (mod 4), then x^2 ≡ −1 (mod p) has a
solution.

Use Euler's Criterion to determine whether
a is a quadratic residue mod
p:
a. a = 2 p = 17
b. a = 7 p = 13

Let p be an odd prime and let a be an odd integer with p not
divisible by a. Suppose that p = 4a + n2 for some
integer n. Prove that the Legendre symbol (a/p) equals 1.

Let p be an odd prime.
Prove that −1 is a quadratic residue modulo p if p ≡ 1 (mod 4),
and −1 is a quadratic nonresidue modulo p if p ≡ 3 (mod 4).

Use the fact that each prime possesses a primitive root to prove
Wilson’s theorem:
If p is a prime, then (p−1)! ≡ −1 (mod p).

Let gcd(m1,m2) = 1. Prove that a ≡ b (mod m1) and a ≡ b (mod m2)
if and only if (meaning prove both ways) a ≡ b (mod m1m2). Hint: If
a | bc and a is relatively prime to to b then a | c.

Let p be prime. Show that the equation x^2 is congruent to 1(mod
p) has just two solutions in Zp (the set of integers). We cannot
use groups.

prove that n> 1 and (n-1)!congruence -1(mod n) then n is
prime

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