If $a$ and $b$ are odd, then $(2a,2b)=(a+b, a-b)$, where $(x,y)$ denotes the greatest common divisor of $x$ and $y$.

Question

I am trying to prove the following assertion:

"If $a$ and $b$ are odd, then $(2a,2b)=(a+b, a-b)$".

$(x,y)$ denotes the greatest common divisor of $x$ and $y$. I am trying to prove it by showing that $(a,b) \vert (a+b, a-b)$ and $(a+b, a-b) \vert (a,b)$.

Denote $(a+b, a-b)$ by $d.$ Since $d \vert (a+b)$ and $d \vert (a-b)$ then $d \vert (a+b) + (a-b) = 2a$ and also $d \vert (a+b)-(a-b) = 2b$. This implies that $d \vert (2a,2b) = 2(a,b)$.

The other way of the proof is where I am struggling. Suppose $(a,b) = y$. Since $a$ and $b$ are odd we have that $a=2n+1$ and $b=2m+1$ for some integers $n,m.$ The sum $a+b$ and the difference $a-b$ is therefore even, and also we have that $a=yr$ and $b=ys$ for some integers $r,s$ so $a+b = y(r+s)$ and $a-b = y(r-s).$ This implies that $y \vert (a+b)$ and $y \vert (a-b)$ which implies that $y \vert (a+b, a-b)$. However, I have not been able to show that $2y \vert (a+b, a-b)$. Any pointers in the right direction would be much appreciated!

Answer 1

Cancelling $\,d = (a,b)\,$ from both sides reduces it to the case where $\,(a,b) = 1.\,\,$ Then it becomes $\, 2 = (a-b,\,a+b)\,$ for $a,b\,$ odd and coprime. Proof: $ $ clearly $\,2\mid a-b,a+b\,$ by $\,a,b\,$ odd. Conversely $\,c\mid a-b,a+b\,\Rightarrow\, c\mid 2a,2b\,\Rightarrow\, c\mid (2a,2b)=2(a,b)=2.\ $ QED

Alternatively: $\,a\!+\!b,\,a\!-\!b\,$ are divisible by the coprimes $\,2\,$ and $\,(a,b)\,$ so also by their product $2(a,b) = (2a,2b),\,$ so $\,(2a,2b)\mid (a\!+\!b,a\!-\!b).\,$ This completes the proof (with your converse).

Answer 2

Let $D=\gcd(2a,2b)$ and $d=\gcd(a+b,a-b)$

Clearly $2\,|\,D$ so let $D=2E$. As you note, $E=\gcd(a,b)$. It is clear that $2\,|\,d$, and it is clear that $E$ is odd. To see that $E\,|\,d$ we remark that $E$ must divide both $a$ and $b$, hence $E$ divides $a\pm b$ and we are done.

Answer 3

$ D=(a,b)$ so that $D|a,\ D|b,\ 2D=(2a,2b)$ and $D$ is odd

Hence $D|a+b,\ D|a-b$ so that $D|(a+b,a-b)$

Since $a+b,\ a-b$ are even, $2D|(a+b,a-b)$

When $D'=2kD|(a+b,a-b),\ k>1$, then $$2kD=D'|(a+b)-(a-b)=2b ,\ kD|b$$

Similarly, $kD|a$ so that it is a contradiction to $D=(a,b)$