Question

Show if p >1 and p divides (p -1)! + 1, then p is prime.

I need a direct proof.

Answer #1

suppose p is a prime number and p2 divides ab and gcd(a,b)=1.
Show p2 divides a or p2 divides b.

A natural number p is a prime number provided that the only
integers dividing
p are 1 and p itself. In fact, for p to be a prime number, it is
the same as requiring that
“For all integers x and y, if p divides xy, then p divides x or p
divides y.”
Use this property to show that
“If p is a prime number, then √p is an irrational number.”
Please write down a formal proof.

Show that if n2+2 and n2-2 are
both prime, then 3 divides n.
Note: This question was asked in the context of congruences/
modulus in a Number Theory class.
Please give a NEATLY written proof in full sentences. State any
theorems, definitions, and formulas used.

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.

Suppose p is a positive prime integer and k is an integer
satisfying 1 ≤ k ≤ p − 1. Prove that p divides p!/ (k! (p-k)!).

Two numbers are relatively prime if their greatest
common divisor is 1. Show that if a and b are relatively prime,
then there exist integers m and n such that am+bn = 1. (proof by
induction preferred)

Formal Proof: Let p be a prime and let a be an
integer. Assume p ∤ a. Prove gcd(a, p) = 1.

: (a) Let p be a prime, and let G be a finite Abelian group.
Show that Gp = {x ∈ G | |x| is a power of p} is a subgroup of G.
(For the identity, remember that 1 = p 0 is a power of p.) (b) Let
p1, . . . , pn be pair-wise distinct primes, and let G be an
Abelian group. Show that Gp1 , . . . , Gpn form direct sum in...

Prove that for any integer a, k and prime p, the following three
statements are all equivalent: p divides a, p divides a^k, and p^k
divides a^k.

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).

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