There are 2 cases. 1. $a=b$. Then $x=a+1$, $y=a$ is a solution. 2. $a \ne b$. WLOG we can assume that $a>b$. Then there is integer $c$ such that $b< c \le a$ and $a-b \mid c$. Let $c=k(a-b)$. Then $x=k(2(c-b)-1)$, $y=k(2(a-c)+1)$ is a solituon.

Same topic was discussed there. http://math.stackexchange.com/questions/804692/existence-of-x-y-satisfying-diophantine-equation/807459#807459 And wrote a formula for the solution. These brackets denote? Rewrite this equation in normal form, so that it was clear and there was no clear understanding.

These parentheses on the LHS mean a binomial coefficient. How many ways can you choose two things from $x+y$?

Possibly a motivated solution: WLOG $a \geq b$. If $a=b$, we simply set $x=2a,y=1$ and it is easy to check that conditions are satisfied. Now comes $a>b$. Let $z=x+y$ so that condition rewrites as: \[\frac{z(z-1)}{2}=bz+(a-b)x\]\[\frac{z(z-1-2b)}{2}=(a-b)x\]\[z(z-1-2b)=(2a-2b)x\]Above implies that there is a positive $k \in \mathbb{Q^{+}}$, which we'll manually choose later, such that: $k=\frac{z}{2a-2b}=\frac{x}{z-2b-1}$, which gives solution $(x,z)=(k(2k(a-b)-2b-1),2k(a-b))$ Observe that $x>0$, so $k>\frac{2b+1}{2a-2b}$ must be satisfied. On the other hand, $z>x$, which rearranges to: $k(2k(a-b)-2b-1)<2k(a-b)\\ 2k(a-b)-2(a-b)<2b+1\\ (2a-2b)(k-1)<2b+1\\ k<\frac{2b+1}{2a-2b}+1$ Now, we choose $k=\left \lceil \frac{2b+1}{2a-2b} \right \rceil$. Observe that $k$ satisfies both properties because $\frac{2b+1}{2a-2b}$ is not integer. Also, both $x$ and $y$ are now positive integers for obvious reasons, and the proof is completed