Is this enough to prove that the GCD is larger?

Question

Prove that $(a+b, a-b) \geq (a, b)$

My attempt

Let $(a+b, a-b) = d$ and $(a, b) = c$. Since $c \mid a,b$

$c$ is also a factor of $a+b$ and $a-b$. Thus $c \leq d$.

Is this enough as a proof? It feels kind of skimpy to me, so I was hoping for some feedback and ideas to make it more rigorous from the MSE community. In particular, I feel as though it's not rigorous enough on showing that $c$ is not always equal to $d$ and thus put some actual meaning behind the 'less than' part of $\leq$.

Answer 1

Your proof seems to be OK, as every number that divides two numbers, also divides their greatest common divisor. And it's well known that $c \mid d \implies c\le d$, when working in positive integers.

For your second requirement here's how one can generate the cases when the GCDs are not equal. By Euclid's Algorithm we have $(a+b,a-b) = (a+b,2b)$ Now obviously $c \mid (a+b,2b)$ and let $a=ca', b=cb'$ where $(a',b')=1$. Then we have that: $(a+b),2b) = c\cdot(a'+b',2b')$ If a factor divides $b'$ then it can't divide $a'+b'$, as $(a',b') = 1$. So $(a'+b',2b')=1$ or $2$. Hence we have that $d=c$ or $2c$. The second case is obtained when $a=ca', b=cb'$, where $a',b'$ are odd coprime integers. While in every other case $d=c$.

Answer 2

Your proof is fine.

Here is a more concise proof:

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

$(a, b) = (a+b,b)=(b,a+b)$

It is easy to show that $(2x,y) \geq (x,y)$

Therefore $(a+b, a-b)=(2b,a+b)\geq(b,a+b)=(a, b) $

Answer 3

Your proof is correct, it shows that c divides d, so c may be less than or equal to d. To see the case of equality, take $a=5$ and $b=2$ then $(5,2)= (7,3)=1 $ and we're done.