Existence of $x,y$ Satisfying Diophantine Equation Let $a,b$ be positive integers. Prove that there exist positive integers $x,y$ such that
$$ \dbinom{x+y}{2} = ax+by $$
 A: First, we will find integer solutions. Clearly, we have a degree 2 polynomial, and it is in fact a conic. The equation is
$$ (x+y)(x+y-1) = 2ax + 2by$$
This seems ugly to deal with, but you should recognize (through discriminants) that it is a quadratic curve. We do a standard change of variables to make it easier to work with. Set $ s = x+y$ and $ t = x-y$, we get that
$$ s^2 - s = a( s+t) + b ( s- t) \Rightarrow (a-b) t = s^2 - (1+a+b) s. $$
Remember that in order to get integer solutions of $x, y$, we need to ensure that $s, t$ have the same parity. We'd simply force both numbers to be even.
Since $(a-b)$ divides the LHS, an easy family of solutions to guess is when $ s = 2 (a-b) n$,where $n$ is an integer. This gives us $ t = 2(a-b)n^2 - (1+a+b) * 2n$. Hence, there are infinitely many integer solutions, of the form
$$ x = (a-b)n^2 + (-2b-1) n , y = - (a-b) n^2 + (1+2a)n.$$
Second, we will find positive solutions. WLOG, we will assume that $ a \geq b$.
To ensure that $x $ is positive, we require that $n > 0$ or $ n < - \frac{ 1+2b}{a-b}$.
To ensure that $y$ is positive, we require that $ - \frac{1+2a}{a-b} < n < 0 $.
It remains to show that there is an integer in the range $ - \frac{1+2a}{a-b} < n < - \frac{1+2b} { a-b}$. But this is obvious, because the range has length 2, hence has an integer (that is not the endpoints).
Specifically, set $n =  - \lfloor \frac{ 1+2b} { a-b} + \epsilon \rfloor $
Note: I did not classify all possible integer solutions. It just happens that this works nicely. From the quadratic equation, you can see that there are finitely many positive integer solutions (first quadrant is finite). We were lucky that this approach worked.
A: Call $\dbinom{x+y}{2}=C$ and treat it as a dummy in a normal Diophantine $C=ax+by$.
The Euclidean algorithm for the GCD(a,b)=d guarantees that there exist x and y such that:
$d=ax+by$,
so whatever $C$ may be, it better be that $d|C$, otherwise the Diophantine is not solvable. But then, rewrite the equation as:
$C=\frac{C}{d}\cdot d=ax\cdot\frac{C}{d} +by\cdot\frac{C}{d}$,
and the numbers are exactly given as:
$X=x\frac{C}{d}$ and
$Y=y\frac{C}{d}$, 
where d, x and y are all given by the Euclidean algorithm.
Verify with Maple for this particular $C$:

X:=x*(x+y)*(x+y-2)/2/d;
Y:=y*(x+y)*(x+y-2)/2/d;

Now, $d|C$ so both the above X and Y are integers, and further:

factor(expand(a*X+b*Y));

$\frac{(x+y)(x+y-2)}{2d}\cdot (ax+by)=C$
A: Solutions of the equation:  $\frac{(x+y)(x+y-1)}{2}=ax+by$
Can be written without using Pell's equation:
$x=\frac{(a+b+1)(3b-a+1)+2(b-a)k-k^2}{8(b-a)}$
$y=\frac{(a+b+1)(b-3a-1)+2(b-a)k+k^2}{8(b-a)}$
$k$ - what some integer.
