Log in to add a comment

## Answers

The Brainliest Answer!

Lemma 1. Let x and y be (not necessary positive) integers and let n be a positive integer. Given an arbitrary prime p (i.e. we can have p = 2) such that 1 gcd(n, p) = 1, p | x − y and neither x nor y is divisible by p (i.e. p ∤ x and p ∤ y). We have vp(x n − y n ) = vp(x − y).

Lemma 2. Let x and y be (not necessary positive) integers and let n be an odd positive integer. Given an arbitrary prime p (i.e. we can have p = 2) such that gcd(n, p) = 1, p | x + y and neither x nor y is divisible by p

Let p be a prime number and let x and y be two (not necessary positive) integers which are not divisible by p. Then: 6 a) For a positive integer n if • p 6= 2 and p | x − y, then vp(x n − y n ) = vp(x − y) + vp(n). • p = 2 and 4 | x − y, then v2(x n − y n ) = v2(x − y) + v2(n). • p = 2 and 2 | x − y, then v2(x n − y n ) = v2(x − y) + v2(x + y) + v2(n) − 1. b) For an odd positive integer n, if p | x + y, then vp(x n + y n ) = vp(x + y) + vp(n). c) For a positive integer n with gcd(p, n) = 1, we have vp(x n − y n ) = vp(x − y). and if n be odd with gcd(p, n) = 1, then we have vp(x n + y n ) = vp(x + y).

Source;MY POSTS FROM THE AOPS