# Relatively prime proof involving a^n and b^n

• RJLiberator
In summary, the proof shows that if a and b are relatively prime natural numbers, then a^n and b^n are also relatively prime. This is because a^n and b^n have the same prime factorization as a and b, and by the corollary, these numbers must be relatively prime. Therefore, there cannot be a prime number that divides both a^n and b^n when a and b are relatively prime.
RJLiberator
Gold Member

## 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?

RJLiberator said:

## 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

RJLiberator said:
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.
RJLiberator said:
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.

RJLiberator

## 1. What does it mean for two numbers to be relatively prime?

Two numbers are relatively prime if they share no common factors other than 1. In other words, the greatest common divisor (GCD) of the two numbers is 1.

## 2. Can you provide an example of two relatively prime numbers?

Yes, for example, 5 and 7 are relatively prime because their only common factor is 1. The GCD of 5 and 7 is 1.

## 3. How can we prove that a^n and b^n are relatively prime?

We can prove that a^n and b^n are relatively prime by showing that their GCD is 1. One way to do this is by using the fundamental theorem of arithmetic, which states that every positive integer can be expressed as a unique product of prime numbers. If the prime factorization of a^n and b^n do not have any common factors, then their GCD will be 1 and they are relatively prime.

## 4. Are a and b also relatively prime if a^n and b^n are relatively prime?

Not necessarily. Just because a^n and b^n are relatively prime, it does not mean that a and b are also relatively prime. For example, a = 4 and b = 9 are not relatively prime, but their fourth powers (a^4 = 256 and b^4 = 6561) are relatively prime.

## 5. Can we generalize this proof to any power greater than n?

Yes, we can generalize this proof to any power greater than n. As long as the prime factorization of a and b do not have any common factors, then their GCD will be 1 and they will be relatively prime. This applies to all powers, not just n.

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