Two problems on gcd I'm almost done with the following two problems, but I need to connect the last few dots.
1:
Let $a \in \mathbb{Z}$ with $a > 0$. Find $(a,a+2)$ and $(3a+5,7a+12)$.
We know that $(a,a+2)=\min\{ma+n(a+2):m,n \in \mathbb{Z}, ma+n(a+2)>0\}$. Choosing $m=-1$ and $n=1$ gives $(a,a+2)=2$. But this clearly only holds when $a$ is even, why doesn't this hold when $a$ is odd? When $a$ is odd, we must have $(a,a+2)=1$.
Similarly, $(3a+5,7a+12)=(-7)(3a+5)+(3)(7a+12)=-21a-35+21a+36=1$, right?
2:
Let $(a,b)=1$ for $a,b \in \mathbb{Z}$. Prove that $(a+b,a-b)$ is either $1$ or $2$.
We know that $ma+nb=1$ for some $m,n \in \mathbb{Z}$. Using the same $m$ and $n$, we have $(a+b,a-b)=ma+mb+na-nb=(ma-nb)+(na+mb)$. By simply switching sign of $n$, we get $ma-nb=1$, but what about the last parenthesis? Why must that be $0$ or $1$? Or is this strategy way off.
Hope you can help.
 A: If you find integers $x$ and $y$ such that $xs+yt=e$, where $e\gt 0$, it can be concluded that $(s,t)\le e$. (Actually, more strongly it can be concluded that $(s,t)$ divides $e$.) 
From $ma+nb=1$, in your first example, you saw that if $m=-1$ and $n=1$, then $ma +n(a+2)=2$. This shows that $(a,a+2)\le 2$. For completeness, one should note that both $(a,a+2)=1$ and $(a,a+2)=2$ are possible, the first when $a$ is odd, and the second when $a$ is even. 
In the second example, you expressed $1$ explicitly as a linear combination of $3a+5$ and $7a+12$. So the gcd of $3a+5$ and $7a+12$ is $\le 1$. Obviously no smaller positive integer can be expressed as a linear combination of $3a+5$ and $7a+12$, so we can conclude that $(3a+5,7a+12)=1$. 
In problem 2,  different letters might be useful. We have $am+bn=1$ for some $m$ and $n$. Therefore we have $m'a+n'(-b)=1$ for some easily found $m'$ and $n'$. It follows that $2$ is a linear combination of $a+b$ and $a-b$. That only shows that the gcd is $\le 2$. We now have to examine cases to see whether $1$ and $2$ are each possible (they are).
Remark: The Bezout Theorem is a useful tool, but it is by no means the only tool. For the second problem, note that if $d$ divides $a+b$ and $a-b$, the $d$ divides their sum and difference, so $d$ divides $2a$ and $2b$. But $d$ must have no common prime factor with $a$, else  $a$ and $b$ would not be relatively prime. So $d$ divides $2$.
A: *

*Obviously, $\gcd(a,b)=\min\{ma+nb|ma+nb>0\}$ is wrong. We say gcd can be represented by that but not the least positive of that.

*Let $c=(a+b,a-b)$, then $c|2a$ and $c|2b$. Since $(a,b)=1$, either $a$ or $b$ is odd, say $a$, namely $(2,a)=1$. Then $c|2a$ implies $c|2$ which means $c$ is 1 or 2, or $c|a$ which means $c|b$ indicating $c=1$. Thus $c$ is either 1 or 2.

