Property of addition, multiplication of $\gcd$ for co-prime $(a,b)=1$.

Question

Need the use of property of addition and subtraction of $\gcd$ for my approach to prove :
$(a,b)=1\implies (ab, a+b)=1$

My approach:
$(a,b)=1\implies (ab,b^2)=b$
$(a,b)=1\implies (ab,a^2)=a$

$((ab,b^2)=b\wedge (ab,a^2)=a)\implies (2ab,a^2+b^2)=b+a\implies (2ab,a^2+b^2+2ab) = a+b \implies (2ab,(a+b)^2) = a+b$

Need help to proceed further, if correct.

Answer 1

Hint :

• if $(a,b)=1$, then $(c, ab) = (c,a)\times (c,b)$
• $(a,b) = (a+b, b)$

Note that you can't "add" gcds so easily. E.g : $(2,6) = 2$ and $(1,4) = 1$ but $(3, 10) = 1$