<s>[INST] 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. 
 [/INST] Consider the equation $f(x,y)=x^2+2xy+y^2-(2a+1)x-(2b+1)y=0$.
If $a=b$ then this is satisfied along the parallel lines $x+y=0$ and $x=y+2a+1$, so we can choose say $x=1$, $y=2a$.
From now on we assume $a>b$. Then the equation is satisfied along a parabola, passing through points $O=(0,0)$, $A=(2a+1,0)$ and $B=(0,2b+1)$ and having its vertex in the second quarter plane.
Starting from the origin, we can perform a sequence of Viéte jumps:
$$
  (x,y) 
  \mapsto \big(x,2b+1-2x-y\big)
  \mapsto \big(2a+1-2(2b+1-2x-y)-x,2b+1-2x-y\big)
  \mapsto \ldots 
$$
All these lattice points lie on the parabola. We continue this sequence until we get onto the arc $AB$.

More formally, let $P_n=(x_n,y_n)$ with
$$
  x_0=y_0=0, \quad
  x_{n+1}=2a-4b-1+3x_n+2y_n,
  \quad y_{n+1}=2b+1-2x_n-y_n.
$$
Notice that $x_{n+1}+y_{n+1}=x_n+y_n+2(a-b)$, so $x_n+y_n=2(a-b)n$. Hence, we must take the unique index $n$ with $2b+1<2(a-b)n<2a+1$ in order to find $P_n$ inside arc $AB$. Solving the equations $f(x,y)=0$, $x+y=2(a-b)n$, the coordinates are
$$
x_n = \big(2(a-b)n-(2b+1)\big)n, \quad
y_n = \big((2a+1)-2(a-b)n\big)n.
$$
(The picture shows the case $a=8$, $b=6$.)
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