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.

Answer 1

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 $4$.

Answer 2

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