Lifting the Exponent (LTE)

The Lifting The Exponent lemma is a powerful tool for finding $v_{p} (x^{n} - y^{n})$ . It quantifies how much "more" divisible by $p$ an expression becomes when we raise terms to a power.

The Lemma (LTE)

Theorem (LTE)

Let

p

be a prime,

n

a positive integer, and

x, y

integers such that:

$p ∣ x - y$
$p ∤ n$
$p ∤ x, p ∤ y$
If $p = 2$ : $n$ is even? No, if $p = 2$ , we usually need $v_{2} (x - y) \geq 1$ .

Actually: Standard LTE requires

p

is odd. Standard Form (Odd $p$ ):

v_{p} (x^{n} - y^{n}) = v_{p} (x - y) + v_{p} (n)

The Case $p = 2$

x, y

are odd and

n

is even:

v_{2} (x^{n} - y^{n}) = v_{2} (x - y) + v_{2} (x + y) + v_{2} (n) - 1

Condition Checklist

Always verify:

Is $p$ odd?
Does $p$ divide the base difference ( $x - y$ )?
Are $x, y$ coprime to $p$ ?

Practice Problems

Exercise (Problem 1)

Find the largest integer

k

such that

3^{k}

divides

202 3^{2023} - 1

? (Wait,

3 ∤ 2022

). Better: Find

v_{3} (5^{2023} - 2^{2023})

Exercise (Problem 2)

Let

p

be an odd prime. Prove that

v_{p} (2^{p - 1} - 1) \geq 1

? No, use LTE to calculate

v_{p} ((p + 1)^{p} - 1)

Exercise (Problem 3)

Find all solutions to

3^{n} = 2^{m} + 1

Lifting the Exponent (LTE)

The Lemma (LTE)

The Case p=2

Condition Checklist

Practice Problems

The Case $p = 2$