- #1

RJLiberator

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## Homework Statement

Show that if a, b, n, m are Natural Numbers such that a and b are relatively prime, then a^n and b^n are relatively prime.

## Homework Equations

Relatively prime means 1 = am + bn where a and b are relatively prime. gcd(a,b) = 1

We have a couple corollaries that may be beneficial:

1. Suppose a, n are positive integers. Then a and a^n have the same prime factors.

2. Let a, n be positive integers. a^(1/n) is either an integer or it's irrational.

## The Attempt at a Solution

By definition of relatively prime, 1 = ax + by where x,y exist as integers.

By the fundamental theorem o arithmetic:

a = p1*p2*...*pm where this is the unique prime factorization.

b = p1*p2*...*pd

m,d are natural numbers.

By the corollarly, a^n = p1^n*p2^n*...*pm^n

b^n = p1^n*p2^n*...*pd^n

So we start with

1 = ax + by

1 = (p1*p2*...*pm)*x + (p1*p2*...*pd)y****All I have left to do is find the way to raise this to the power of n. If I can raise it such that

1^n = (ax)^n + (by)^n the proof is done, easily, with what I have set up.

**Question:**Is this step needed? Can I just use the corollary to say that "by the corollary" since a and a^n have the same prime factorization and b and b^n have the same prime factorization, then we see a^n and b^n must be relatively prime?