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?

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 $d$.
