How to check if p is prime or composite?

How to check if p is prime or composite?

It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is composite, then its positive divisors are among the integers and it is clear that gcd ( ( p -1)!, p) > 1, so we can not have ( p -1)! ≡ -1 (mod p ). However if p is prime, then each of the above integers are relatively prime to p.

What are the inverse inverses of modulo a prime?

Inverses, Modulo a Prime. Theorem 1 When n is a prime number then it is valid to divide by any non-zero number — that is, for each a ∈ {1,2,…,n−1} there is one, and only one, number u ∈ {1,2,…,n−1} such that au = 1 (mod n).

How do you find the prime number with P-1?

p is prime if and only if (p-1)! = -1 (mod p). This beautiful result is of mostly theoretical value because it is relatively difficult to calculate (p-1)! In contrast it is easy to calculate a p-1, so elementary primality tests are built using Fermat’s Little Theorem rather than Wilson’s.

How do you determine if an integer is prime?

Let p be an integer greater than one. p is prime if and only if ( p -1)! ≡ -1 (mod p ). This beautiful result is of mostly theoretical value because it is relatively difficult to calculate ( p -1)! In contrast it is easy to calculate ap-1, so elementary primality tests are built using Fermat’s Little Theorem rather than Wilson’s.

How do you find the proof of p -1?

Proof. Start by listing the first p -1 positive multiples of a: Suppose that ra and sa are the same modulo p, then we have r = s (mod p ), so the p -1 multiples of a above are distinct and nonzero; that is, they must be congruent to 1, 2, 3., p -1 in some order. Multiply all these congruences together and we find

What is the prime divisors theorem?

Prime Divisors Theorem 1. If n>1 is composite, then nhas a prime divisor psuch that p2 . Remark. Another way to say this is that a composite integer n>1 has a prime divisor p with p p n. So if an integers n>1 is not divisible by any prime p p n, we can conclude that nmust be a prime.

How do you use Fermat’s little theorem to find mod p?

Fermat’s little theorem gives you x 2 ≡ 1 ( mod p). Now use the rest of the information you’ve been given. For the first part, note that x p − 1 ≡ 1 by Fermat’s little theorem. Then assuming that p is not 2 (deal with that case separetly), we have 1 ≡ ( x 2) ( p − 1) / 2 ≡ a ( p − 1) / 2.

What is a(p – 1) / 2 = – 1 mod p?

Hence a ( p − 1) / 2 = − 1 mod p. Among the many proofs of these facts, let me mention one that uses basically only Fermat’s little theorem, and then the fact that a polynomial of degree n over a field has at most n roots. And then of course you need to know that Z / p Z is a field, for p prime.

What is the prime factorization of 4 K + 3?

(10.) Prove that any positive integer of the form 4 k + 3 must have a prime factor of the same form. Because 4 k + 3 = 2 ( 2 k + 1) + 1, any number of the form 4 k + 3 must be odd. — so they must have forms 4 k + 1 and 4 k + 3.

How do you prove that there are infinitely many primes?

Theorem 1. There are infinitely many primes. Proof. Suppose there are only finitely many primes. Call these p1 , p2 ., pn . Form the number: N = (p1 p2 …pn ) + 1. This number is certainly bigger than 1 so must have at least one prime factor. Let p be one of these. Now p is prime.