Question

Let m = 2k + 1 be an odd integer. Prove that k + 1 is the multiplicative inverse of 2, mod m.

Answer #1

If p = 2k − 1 is prime, show that k is an odd integer or k =
2.
Hint: Use the difference of squares 22m − 1 = (2m − 1)(2m +
1).

1. Prove that {2k+1: k ∈ Z}={2k+3 : k ∈ Z}
2. Prove/disprove: if p and q are prime numbers and p < q,
then 2p + q^2 is odd (Hint: all prime numbers greater than 2 are
odd)

1. Let n be an integer. Prove that n2 + 4n is odd if and only if
n is odd? PROVE
2. Use a table to express the value of the Boolean function x(z
+ yz).

Suppose that a is an odd integer and (a,91)=1. Prove that
a^12≡1(mod 1456).

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 n be a positive odd integer, prove gcd(3n, 3n+16) = 1.

Let n be an odd integer.
Prove that 5460 | n25 −n

3.a) Let n be an integer. Prove that if n is odd, then
(n^2) is also odd.
3.b) Let x and y be integers. Prove that if x is even and y is
divisible by 3, then the product xy is divisible by 6.
3.c) Let a and b be real numbers. Prove that if 0 < b < a,
then (a^2) − ab > 0.

let n be an odd integer ,prove that 5460 | n^25-n

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.

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