# GCD and divisibility

Update: I think there was a typo in the text. Please don't waste your time with this problem. :)

If gcd$(a,b) = p$, a prime, then $p|am$ and $p|an$ such that gcd$(m,n) = 1$

Why does gcd$(m,n)$ have to be equal to 1? And does this statement hold true if gcd$(a,b)$ is not equal to a prime?

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Let $a=7$ and $b=14$, then $\gcd(a,b) = 7$ which is prime. Obviously $7|7m$ and $7|7n$ for any positive integers $m$ and $n$. Let $m = 2$ and $n = 4$ then $7|7\times 2$ and $7|7\times 4$, yet $\gcd(2,4) = 2 \neq 1$.
Your original statement seems like it is miscopied from somewhere, or missing additional context. If $\gcd(a,b)=p$, prime or not, then $a=pc$ and $b=pd$, for some integers $c,d$ such that $\gcd(c,d)=1$.
The other answer points out that if you don't have some sort of conditions on $m,n$ the original statement is incorrect. You will note that both answers suggest this possibility. – vadim123 May 11 '13 at 23:10