Assuming that $(a, b) = 2$, prove that $(a + b, a − b) = 1$ or $2$

Question

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.

@FredrikMeyer For a moment there, I read that as 4 factorial. Sure threw me off! — ladaghini, Aug 04 '12 at 09:16
I believe what the statement meant to say was that if $(a, b) = 2$, then prove that $2 | (a+b, a-b)$. — ladaghini, Aug 04 '12 at 09:26
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 04 '12 at 10:27
A correct result would be if $(a,b)=2$ then $(a+b,a-b)=2$ or $4$. And both are possible. Certainly there is a typo. — André Nicolas, Aug 04 '12 at 11:47
For the statement from Gerry Myerson comment see: Prove $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ if $\gcd(a,b) = 1$ — Martin Sleziak, Jul 27 '17 at 13:23
If you assume assume that $\operatorname{gcd}(a,b)=1$, then you have $\operatorname{gcd}(a+b, a-b)=1$ or $2.$ — , Apr 11 '20 at 23:20

score 2 · Accepted Answer · edited Jul 27 '17 at 13:22

2

Suppose the $d=\gcd(a+b,a-b)$. Then $d\mid2a$, $d\mid2b$ since $2a=(a+b)+(a-b)$ and $2b=(a+b)-(a-b)$. Then $d\mid\gcd(2a,2b)=2\gcd(a,b)=4$. Since $a+b$, $a-b$ are even then $d$ is $2$ or $4$.

edited Jul 27 '17 at 13:22

Martin Sleziak

53,687

answered Aug 04 '12 at 12:41

i. m. soloveichik

4,259

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 04 '12 at 15:07
@labbhattacharjee Sorry I missed that. I fixed it. – i. m. soloveichik Aug 04 '12 at 15:18

score -1 · Answer 2 · edited Jul 27 '17 at 13:22

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

$\Rightarrow d\mid(a+b) \pm (a-b) \Rightarrow d\mid2a$ and $d\mid2b \Rightarrow d\mid(2a,2b) \Rightarrow d\mid2(a,b)$

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 $P$, $Q$, $(a,b)\mid(P.a+Q.b)$, $d$ 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).

Observation:

$a$,$b$ 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.$(6, 10)=2 \Rightarrow (6+10, 6 - 10)=4$

Assuming that $(a, b) = 2$, prove that $(a + b, a − b) = 1$ or $2$

2 Answers2