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Proof question.
I am asked to show $\gcd(a,b)=1$ implies that $\gcd(a+b,a-b)\leq 2$. I am not sure if my solution is correct.

Suppose $\gcd(a,b)=1$. Then, there are integers, x and y, such that $ax+by=1$. We claim there are integers $s$ and $t$ such that $(a+b)s+(a-b)t=2$. We can re-arrange this equation and obtain $a(s+t)+b(s-t)=2$.
Now, set $s=x+y$, and $t=x-y$, and we have that $a(x+y+x-y)+b(x+y-x+y) = 2ax + 2by = 2(ax+by)= 2.$
This implies that $\gcd(a+b,a-b)\vert 2$ and so $\gcd(a+b,a-b)\leq 2$

The part I am a little fuzzy on is whether or not setting $s=x+y$, and $t= x- y$ legitimately leads to a general solution. The idea was to obtain a particular solution to the equation, showing that for any $a,b$ we can construct an equation of the form $(a+b)s+(a-b)t=2$, but I am not sure if this is a move I am able to make here, it felt a bit like cheating to me.

Nikolas Koutroulakis
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  • Did you compare with other (standard) solutions, e.g., here?: Let $d \mid a+b, a-b$, then $d \mid (a+b)+(a-b) = 2a$ and $d \mid (a+b)-(a-b) = 2b$ so $d \mid \gcd(2a,2b) = 2$. No need of Bezout's Lemma, with integers $s,t$ etc. – Dietrich Burde Mar 05 '23 at 20:05
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    That's fine, if a bit wordy. You don't need to tell us how you solved it, just solve $(a+b)s+(a-b)t=2$ from $ax+by=1.$ – Thomas Andrews Mar 05 '23 at 20:07
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    This method is one of the standard methods already presented in the linked dupe. Compare your solution to that and then if you have any questions post a question that explains precisely what you doubt, and why so. [For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine]. – Bill Dubuque Mar 05 '23 at 20:11
  • @ThomasAndrews, thanks, it's a bit wordier than I usually right because I wanted to be a bit more explicit about how I was proving it, but if you think I m overexplaining it, I can certainly take that into account for my future posts – Nikolas Koutroulakis Mar 05 '23 at 21:09
  • @DietrichBurde, thank you for directing me here! – Nikolas Koutroulakis Mar 05 '23 at 21:13

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