Q:

Does ax+by = (a+b)z has a solution?

For any pair of integers x and y, can we always find two integers a and b such that the equation $ax+by=(a+b)z$, $z$$\in$ $Z$

A:

Yes, it can always be done.

Let $x,y$ be given integers.

Then choosing any integer $b$, and letting $a=-2b$, we have $a+b=-b$, hence
\begin{align*}
&ax + by
\qquad\qquad\qquad\;\;
\\[4pt]
=\;\,&-2bx + by\\[4pt]
=\;\,&-b(2x-y)\\[4pt]
=\;\,&(a+b)(2x-y)\\[4pt]
\end{align*}
More generally, if $t,k$ are any integers, then letting
\begin{align*}
a&=kt
\qquad\qquad\qquad
\\[4pt]
b&=(1-k)t\\[4pt]
\end{align*}
we have $a+b=t$, hence
\begin{align*}
&ax + by\\[4pt]
=\;\,&(kt)x + (1-k)ty\\[4pt]
=\;\,&t(kx+(1-k)y)\\[4pt]
=\;\,&(a+b)\bigl(kx+(1-k)y\bigr)\\[4pt]
\end{align*}

