# Elementary Number Theory (GCD)

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.

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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

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.

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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).

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

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