On existence of positive integer solution of $\binom{x+y}{2}=ax+by$ How can I prove this? 

Prove that for any two positive integers $a,b$ there are two positive integers $x,y$ satisfying the following equation:
  $$\binom{x+y}{2}=ax+by$$

My idea was that $\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$ and choose $x,y$, such that $2a=x+2y-1, 2b=y(y-1)$, but using this idea, $x,y$ won’t be always positive. 

If $a=b$ then let $(x,y)=(a,a+1)$. 
Otherwise, w.l.g. suppose $a>b$ and let $x+y=2t(a-b)$ for some positive integer $t$. Then 
$$t(a-b)\Big(2t(a-b)-1\Big) =ax+by=(a-b)x+2bt(a-b)$$
$\text{Therefore } x=t\Big(2t(a-b)-1\Big)-2bt=t\Big(2t(a-b)-(2b+1)\Big)$.
$x$ will be a positive integer providing $t>\frac{2b+1}{2(a-b)}.$
$y=2t(a-b)-x$ will be a positive integer providing $2(a-b)>2t(a-b)-(2b+1)$ i.e. $\frac{2b+1}{2(a-b)}>t-1$.
$\frac{2b+1}{2(a-b)}$ is positive but not an integer and so, precisely as required, there is a positive integer $t$ such that 
$$t>\frac{2b+1}{2(a-b)}>t-1.$$
