Proof that if $\gcd(m,n) = 1$, then $\gcd(m+n,mn ) = 1$. 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.
 A: $(a+b)^2-(a-b)^2=4ab$ 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.
A: 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.
A: 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 $mn$. 
