# Number Theory(a,b)=1 => (a^m,b^n)=1

#### evinda

##### Well-known member
MHB Site Helper
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
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 $b^n$...