Prove $\gcd(a,b)=\gcd(a+b,\operatorname{lcm}(a,b))$

Question

Any ideas on how to prove this equality? I tried various methods, using properties of gcd and lcd, but I can't prove it.

Answer 1

First we prove that if $\gcd(u,v)=1$ then $\gcd(u+v,uv)=1$

Proof:

If $\gcd(u+v,uv)\neq 1$ there is a prime $p$ such is $p|u+v$ and $p|uv$

Now if $p|uv$ then $p|u$ or $p|v$. Without loss of generality we can assume $p|u$.

Now from $p|u+v$ and $p|u$ we get $p|v$. So $p$ is a common factor of $u$ and $v$. Contradiction.

Going back to the problem:

Assume $d=\gcd(a,b)$ we have $a=du$ and $b=dv$ with $\gcd(u,v)=1$

Also we have $\text{lcm}(a,b)=duv$

So on the right side we have:

$\gcd(d(u+v),duv)=d\gcd(u+v,uv)=d$

Which was to be proved.

Answer 2

You may want to try and prove first that

$\operatorname{gcd}(a, b) = \operatorname{gcd}(b, a - b)$

This identity should be enough to get you rolling. You will just have to be able to use it correctly.

Alternatively, try this approach. Suppose $d $ is the gcd of $a $ and $b $. Show it divides $a+b $ and $\operatorname {lcm}(a, b) $.

Then suppose $d'$ is the gcd of $a+b $ and $\operatorname {lcm}(a, b) $. Show $d'$ divides both $a $ and $b $. Hence $d $ divides $d'$ which divides $d $, meaning they must be equal.

Answer 3

Let gcd(a,b)=d. a=ds , b=de . a+b=d(e+s). lcm(a,b)=ab/d=des. Assume x greater than d divides des,d(e+s). By the first it divides one of e and s. But if it divides one, it must divide both by the second. But hten dx is the gcd a contradiction.