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

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