Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Statement to be proved: Assuming that $(a, b) = 2$, prove that $(a + b, a − b) = 1$ or $2$.

I was thinking that $(a,b)=\gcd(a,b)$ and tried to prove the statement above, only to realise that it is not true.

$(6,2)=2$ but $(8,4)=4$, seemingly contradicting the statement to be proved?

Is there any other meaning for $(a,b)$, or is there a typo in the question?

Sincere thanks for any help.

share|cite|improve this question
It is true that (8,4)=4! – Fredrik Meyer Aug 4 '12 at 9:15
@FredrikMeyer For a moment there, I read that as 4 factorial. Sure threw me off! – ladaghini Aug 4 '12 at 9:16
If $(a,b)=2$ then $(a+b, a^2+b^2)= 1$ or $2$. – Quixotic Aug 4 '12 at 9:18
@Quixotic How about (6,2)=2 but (8,40)=8 ? – yoyostein Aug 4 '12 at 9:20
Perhaps the statement to be proved was actually, assuming $\gcd(a,b)=1$, then $\gcd(a+b,a-b)$ is 1 or 2. – Gerry Myerson Aug 4 '12 at 10:27
up vote 2 down vote accepted

Suppose the $d=\gcd(a+b,a-b)$. Then $d|2a$, $d|2b$ since $2a=(a+b)+(a-b)$ and $2b=(a+b)-(a-b)$. Then $d|\gcd(2a,2b)=2\gcd(a,b)=4$. Since $a+b,a-b$ are even then $d$ is 2 or 4.

share|cite|improve this answer
Is d=gcd(a+b, a-b)? Then, how can d be 1? d needs to be even as both a,b are even as ladaghini has noticed in the comment of the question. – lab bhattacharjee Aug 4 '12 at 15:07
@labbhattacharjee Sorry I missed that. I fixed it. – i. m. soloveichik Aug 4 '12 at 15:18

if $d|(a+b, a-b) \Rightarrow d|(a+b)$ and $d|(a-b)$

$\Rightarrow d|(a+b) \pm (a-b) \Rightarrow d|2a$ and $d|2b \Rightarrow d|(2a,2b) \Rightarrow d|2(a,b) $

This is true for any common divisor of (a+b) and (a-b).

If d becomes (a+b, a-b), (a,b)|(a±b)

as for any integers $P$, $Q$, $(a,b)|(P.a+Q.b)$, $d$ can not be less than $(a,b)$,

in fact, (a,b) must divide d,

so $d = (a,b)$ or $d = 2(a,b)$.

Here (a,b)=2.

So, $(a+b, a-b)$ must divide $4$ i.e., $= 2$ or $4$ (as it can not be $1$ as $a$, $b$ are even).


$a$,$b$ can be of the form

(i) $4n+2$, $4m$ where $(2n+1,m)=1$, then $(a+b, a-b)=2$, ex.$(6,4) = 2 \Rightarrow (6+4, 6-4)=2$

or (ii) $4n+2$, $4m+2$ where $(2n+1,2m+1)=1$, then $(a+b, a-b)=4$, ex.$(6, 10)=2 \Rightarrow (6+10, 6 - 10)=4$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.