Prove $(a, b) \mid ((a + b), (a - b))$ I tried this:
Suppose $(a, b) = d$. Then $ax + by = d$. 
Let $((a + b), (a – b)) = e$. 
Then $$\begin{align}e& = (a + b)u + (a – b)v\\
&= au + bu + av – bv\\
&= a(u + v) + b(u – v)\end{align}$$
Let $u + v = x$ and $u – v = y$, then $d = e$. So, $d \mid e$. 
What are the possible errors?
 A: There is some slight confusion about what you really proved, but it's minor (even though, rigorously speaking, the proof is all wrong).
You can actually remove the first sentence "Then $ax+by=d$".
The argument would flow better like so:


*

*We know: by Bezout's theorem, if there exists a pair $x,y$ such that $ax+by=z$, then $d|z$

*Suppose $(a,b)=d$ and $((a+b),(a-b))=e$

*Then, we know that there exists a pair $u,v$ such that $e=u(a+b)+v(a-b)$

*Then, by setting $x=u+v, y=u-v$, we see that $ax+by=e$

*We have proven that there exists a pair $x,y$ such that $ax+by=e$, therefore, by point $1$, we know that $d|e$

A: If $(a, b) = d$, then we have
$$a+b = d\cdot\left(\frac{a}{d} + \frac{b}{d}\right)\text{ and }a-b = d\cdot\left(\frac{a}{d} - \frac{b}{d}\right)$$
and obviously then
$$((a+b), (a-b)) = \left(d\cdot\left(\frac{a}{d} + \frac{b}{d}\right), d\cdot\left(\frac{a}{d} - \frac{b}{d}\right)\right) = d\cdot\left(\left(\frac{a}{d} + \frac{b}{d}\right), \left(\frac{a}{d} - \frac{b}{d}\right)\right),$$
which even more obviously is dividable by $d$.
