What is lifting the exponent lemma
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Lifting The Exponent Lemma is a new topic in number theory and is a
method for solving some exponential Diophantine equations. In this article we
analyze this method and present some of its applications.
We can use the Lifting The Exponent Lemma (this is a long name, let’s call
it LTE!) in lots of problems involving exponential equations, especially when
we have some prime numbers (and actually in some cases it “explodes” the
problems). This lemma shows how to find the greatest power of a prime p –
which is often ≥ 3 – that divides a
n ±b
n for some positive integers a and b. The
proofs of theorems and lemmas in this article have nothing difficult and all of
them use elementary mathematics. Understanding the theorems usage and its
meaning is more important to you than remembering its detailed proof.
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
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
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The Lifting the Exponent (LTE) Lemma is a useful lemma about the largest power of a prime dividing a difference or sum of nth powers. it is used in especially prime numbers in lots of problems involving exponential equations.This lemma shows how to find the greatest power of a prime
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