Relatively prime proof involving a^n and b^n

[RJLiberator]

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?

 

[Samy_A]

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?

Yes to the last question. I was wondering why this looks so complicated.
Can there be a prime number that divides both $a^n$ and $b^n$, when a and b are relatively prime?

 

[RJLiberator]

Yes to the last question. I was wondering why this looks so complicated.

Since a and a^n and b and b^n have the same prime factorizations, and a,b are relatively prime, then a^n,b^n are relatively prime.

Can there be a prime number that divides both an" role="presentation" style="display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">anan and bn" role="presentation" style="display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">bnbn, when a and b are relatively prime?

No.
Since they are relatively prime, that means 1 = ax+bm and by this proof, 1 = a^n*x+b^n*m

 

[Samy_A]

Since a and a^n and b and b^n have the same prime factorizations, and a,b are relatively prime, then a^n,b^n are relatively prime.

Correct.

No.
Since they are relatively prime, that means 1 = ax+bm and by this proof, 1 = a^n*x+b^n*m

Hmm, A little sloppy.
The "x" and "m" in the RHS don't have to be the same as the "x" and "m" in the LHS.

Example: 5*2-3*3=1, but 5²*2-3²*3≠1. (But -5²*5+3²*14=1)
It doesn't matter, you already solved the exercise.