MHB Solving $(a,b)=1 \Rightarrow (a^m,b^n)=1$ without Primes

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To prove that if $(a,b)=1$, then $(a^m,b^n)=1$ without using primes, one approach involves using Bézout's identity, which states there exist integers x and y such that $ax + by = 1$. By expanding $(ax + by)^{m+n}$, it can be expressed as a linear combination of $a^m$ and $b^n$. This method avoids the need to show that $(a^{m-1},d)=1$ and $(b^{n-1},d)=1$ directly. The discussion emphasizes finding alternative methods to demonstrate the coprimality of powers without prime factorization. The conversation highlights the importance of exploring different mathematical techniques in proofs.
evinda
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Hey! :)

I have to show that if $(a,b)=1 \Rightarrow (a^m,b^n)=1$,without using primes!
Suppose that $d=(a^m,b^n)$.Then $d|a^m , d|b^n$.
How can I continue?
Do I have to show that $(a^{m-1},d)=1$ and $(b^{n-1},d)=1$? If yes,how could I do this? :confused:
 
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evinda said:
Hey! :)

I have to show that if $(a,b)=1 \Rightarrow (a^m,b^n)=1$,without using primes!
Suppose that $d=(a^m,b^n)$.Then $d|a^m , d|b^n$.
How can I continue?
Do I have to show that $(a^{m-1},d)=1$ and $(b^{n-1},d)=1$? If yes,how could I do this? :confused:

Hi!

I have a different approach.

From $(a,b)=1$ we know that there are numbers x and y such that $ax+by=1$ (Bézout's identity).

Now expand $(ax+by)^{m+n}$ and write it as a linear combination of $a^m$ and $b^n$...
 
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