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

Question

Let $d=(a,b)$. Since $d\mid(a+b)$ and $d\mid(a-b)\Rightarrow d\mid(a+b)+(a-b)=2a$ Similarly, $d\mid2b$. With $a$ and $b$ odd, $d\mid2(2k+1)$ and $d\mid2(2j+1)$.

$$\begin{align}(a+b)+(a-b)&=(2k+1+2j+1+2k+1-2j-1)\\ &=4k+2\\ &=2(2k+1)2a(a+b)-(a-b)\\ &= (2k+1+2j+1-(2k+1-(2j+1)))\\ &= 2(2j+1)=2b\end{align}$$

I feel like I'm going in circles. From here, how can I go to $2(a,b)=(a+b,a-b)$?

Answer 1

Let $d = (a,b)$ and $D = (a+b, a-b)$. We want to show that $2d = D$ (assuming that we take $d, D > 0$).

Since $d \mid a$ and $d \mid b$, it follows that $d \mid a+b$ and $d \mid a-b$, so $d \mid D$. Since $a$ and $b$ are odd, clearly $d$ must be odd too, but since $a+b$ and $a-b$ are even numbers, it follows that $2 \mid D$. Since $2 \nmid d$ it follows that $2d \mid D$.

Conversely, since $D \mid a+b$ and $D \mid a-b$, it follows that $D \mid (a+b) + (a-b) = 2a$ and, similarly, $D \mid 2b$, therefore $D \mid (2a, 2b) = 2 (a,b) = 2d$, so $D \mid 2d$.

Since $2d \mid D$ and $D \mid 2d$ and both numbers are taken to be positive, it follows that $D = 2d$.

Answer 2

You've shown $d$ and $2$ are divisors of $a+b$ and $a-b$. What you need to show is that $2(a,b)$ is the greatest common divisor.

Outline: 1) $2(a,b)$ is a divisor of $a \pm b$.

2) if $e|a\pm b$ then $e|2(a,b)$.

Therefore $2(a,b) = (a+b, a-b)$.

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1) $(a,b)|a \pm b$, $2|a \pm b$ (as $a \pm b$ is even as $a$ and $b$ are odd). As $((a,b),2) = 1$ then $2*(a,b)|a\pm b$.

2)$e|a\pm b$ then $e|2a$ and $e|2b$. If $e$ is odd, $e|a$ and $e|b$ and $e|2(a,b)$. So $e|2(a,b)$ if $e$ is odd.

If $e = 4e'$ then $e = 4e'|2a$ so $2e'|a$ so $a$ is even. A contradiction so $e \ne 4e'$.

If $e = 2e'$ then $e'$ is odd. $2e'|2a$ so $e'|2a$ so $e'|a$ and $e'|b$ and so $e'|(a,b)$ so $e=2e'|2(a,b)$ if $e$ is even.

So $e|2(a,b)$

So $2(a,b)$ is greatest common divisor.