Proof that if $\gcd(m,n) = 1$, then $\gcd(m+n,mn ) = 1$.

Question

I need help with this excercise.

If $\gcd(m,n) = 1$, then $\gcd(m+n,mn ) = 1$.

I don't know how to prove this, I know the definition of $\gcd$ but I can't prove it.

Answer 1

$(a+b)^2-(a-b)^2=4ab$ Therefore if a prime $p$ divides $(a+b)$ and $ab$ it must divides $(a-b)$ from which p divides both $a$ and $b$ which contradicts the hypothesis.

Answer 2

If possible , let us assume that $$gcd(m+n,mn)=d\neq 1$$. Then as a number $d$ must have a prime divisor , say $p$. Since $$d|mn$$ hence $$p|mn$$. As $p$ is a prime number , $p$ must divide at least one of $m$ or $n$. Wlog, let $$p|m$$. Then again $$p|(m+n),\ as \ \ d|(m+n)$$ and $$p|m$$. Thus we have $$p|((m+n)-m)$$ i.e. $$p|n$$.This implies that $$m\ \ and\ \ n\ \ have \ \ a\ \ common\ \ divisor\ \ p\ \ which\ \ is\ \ not\ \ 1\ \ i.e.\ \ gcd(m,n)\neq 1$$. We are at a contradiction,. So our assumption was wrong . $$gcd(mn,m+n)=1$$ must hold.

Answer 3

We know that the prime factors of $mn$ is either a factor of $m$ or $n$ but not the other due to $gcd(m,n) = 1$. However, $m+n$ is not divisible by a prime factor of $n$ nor $m$ because the division yields an integer plus a 'decimal number'. Hence, by the fundamental theorem of arithmetic, it is not divisible by any factors of $mn$.