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Number Theory (a,b)=1 => (a^m,b^n)=1

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,730
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,902
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 $b^n$...