The necessary tools can be found in Glaisher's 1900 paper "Congruences relating to the sums of products of the first n numbers and to other sums of n products", which is more or less available here. (It is the first article in the issue.) Glaisher only proves the m = 2 case this way, I think - but the method works for all cases. Define the polynomial f(x) = (x + 1)(x + 2)⋯(x + p − 1) = [p ...

2015年3月14日 · The multiplicative group mod 2n mod 2 n is isomorphic to Z2n−2 ×Z2 Z 2 n − 2 × Z 2 so we have Z2n−3 Z 2 n − 3 odd quadratic residues. How many even quadratic residues do we have? How many do we have of the form d22k? d 2 2 k? with d d odd? d d is an odd quadratic residue mod 2n−2k mod 2 n − 2 k so there are 2n−2k−3 2 n − 2 k − 3 options. Hence the …

Not really sure how to go about this problem, especially since there is a bi-conditional nested in a larger conditional statement. If anyone could show me how to progress I'd really appreciate it. ...

2012年5月7日 · Assume: p p is a prime that satisfies p ≡ 3 (mod 4) p ≡ 3 (mod 4) Show: x2 ≡ −1 (mod p) x 2 ≡ − 1 (mod p) has no solutions ∀x ∈Z ∀ x ∈ Z. I know this problem has something to do with Fermat's Little Theorem, that ap−1 ≡ 1 (mod p) a p − 1 ≡ 1 (mod p). I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a ...

2010年7月21日 · How can I show that $ (n-1)!\equiv-1 \pmod {n}$ if and only if $n$ is prime? Thanks.

I need the general case, where a a and m m need not to be coprime. Which reduction can I use in this case ? Is there an easier method to calculate the tetration modulo m without using the reductions modulo m, ϕ(m), ϕ(ϕ(m))... m, ϕ (m), ϕ (ϕ (m))... ? I tried to write a program in PARI, but I failed because reducing modulo ϕ(m) ϕ (m) does not work, if a and m are not coprime.

2015年5月9日 · Since a a is not divisible by p p, neither is x x, so of course xp−1 ≡ 1 x p − 1 ≡ 1. We cannot even formulate your assertion when p p is not 3 3 modulo 4 4, since otherwise (p + 1)/4 (p + 1) / 4 is not even an integer. But besides being able to pose a well defined problem, I didn't use p ≡ 3 (mod 4) p ≡ 3 (mod 4) anywhere, only that p p was odd.

If ab a b is congruent to ac (mod n) a c (mod n) and a a is not congruent to 0 (mod n) 0 (mod n), then b b is congruent to c (mod n) c (mod n). This was a homework problem and I was asked to show that this is false for some set of integers. I came up with a = 4, b = 8, c = 5, a = 4, b = 8, c = 5, and n = 6 n = 6. Then 4 ∗ 8 − 4 ∗ 5 = 12 4 ∗ 8 − 4 ∗ 5 = 12, which is divisible by 6 6 ...

I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.

This is probably a really trivial question but I just cannot see the answer... Suppose p p is prime. Then x2 ≡ 1 (mod p) x 2 ≡ 1 (mod p) has only solutions x ≡ ±1 (mod p) x ≡ ± 1 (mod p) for x ∈ Z x ∈ Z. How do I prove this?