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if GCD of $(a, b) = 1$, prove that GCD $(a+b, a-b) = 1$ or $2 .$ The proof goes like: Let GCD $( a+b, a-b ) = d$ and let there exist integers m and n such that $ a+b =md$ and $ a-b = nd.$ By adding and subtracting these two equations we get: $2a = (m+n)d$ and $2b = (m-n)d$ , because $a, b$ are coprime then $2$ GCD $(a,b)$ = GCD $(2a, 2b),$ and so on.

My question is, why do we have to add and subtract above equations? I need to understand the concept of this prove in some more details. Thanks!

Bill Dubuque
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Asim
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  • Not sure I understand the question. You don't have to add those two equations. There's lots of ways to proceed. You could, say mjust remark that $\gcd(a+b,a-b)=\gcd(a+b+a-b, a-b)=\gcd(2a, a-b)=\gcd(2,a-b)$. – lulu Apr 27 '21 at 13:07
  • @lulu i m sure there are other methods but i am particularly trying to understand this method. – Asim Apr 27 '21 at 13:12
  • Well, what don't you understand? They key point is that $d$ must be coprime with both $a$ and $b$...do you see how to show that? – lulu Apr 27 '21 at 13:13
  • See https://math.stackexchange.com/a/1911456/589 – lhf Apr 27 '21 at 13:22
  • Do you understand why if $\gcd(a,b)$ divides $a$ and it divides $b$ then it we divide $a+b$ and it will divide $a-b$. And do you understand and something If $k$ divides $a$ then it will divide $a+b$ if and only if it divides $b$? If you understand that this result is immediate. – fleablood Apr 27 '21 at 22:52
  • We do it because $(a+b)+(a-b) = 2a$ is an expression entirely in terms of $a$ and $(a+b)-(a-b) =2b$ is an expression entirely in terms of $b$. Anything that divides both $(a+b)$ and $(a-b)$ must divide the sum and the difference and that will allow us to conclude any factor that divides both must divide both the expressions entrirly in terms of $a$ or entirely in terms of $b$. – fleablood Apr 27 '21 at 23:01

1 Answers1

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Letting $(x,y)$ denote the gcd of $x$ and $y$, you have $(a,b)=1$ and want to show that $(a+b,a-b)=1 \text{ or }2$. As such, you begin by letting $d=(a+b,a-b)$ and observing that there exist $m,n \in \mathbb{Z}$ such that $$a+b=md \text{ and } a-b = nd.$$ If you add the above equations, you get $2a = (m+n)d$. If you subtract, you get $2b = (m-n)d$. This tells you that $d|2a$ and $d|2b$. As you note, $(2a,2b)=2(a,b)$. This implies that $(2a,2b)=2$. This means that $d=1$ or $d=2$ (since $d$ is a divisor of $2a$ and $2b$, we must have $d\leq(2a,2b)$).

To answer the question about why you add/subtract them, the answer is to isolate $a$ and $b$ on the LHS. This is what allows you to conclude that $d|2a$ and $d|2b$. As mentioned in one of the comments, you don't HAVE TO add/subtract (there are other ways to proceed), but that is why it was done in this case.

Gary Moon
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