\[n=\frac{(x+y)^2+3x+y}{2} =\frac{(x+y)(x+y+1)}{2}+x = \frac{k(k+1)}{2} + x \]= Sum of integers upto \(k\) and the gap between the succeding sum is also \(k\) so \(x \in {0,1,2, , , k}\) can yield that furthermore it's uniqueness is clear.

RC. wrote: \[n=\frac{(x+y)^2+3x+y}{2} =\frac{(x+y)(x+y+1)}{2}+x = \frac{k(k+1)}{2} + x \]= Sum of integers upto \(k\) and the gap between the succeding sum is also \(k\) so \(x \in {0,1,2, , , k}\) can yield that furthermore it's uniqueness is clear.

PS

I think it is wrong. Value of $k$ belong to $x$, so if you let $x_0 \in {0,1,2, , , k}$ different $x$ , you will see it is impossible.

$<Claim 1>$ when $a_n$ is a sequence defined by $a_n=9+8n$ where $n$ is a positive integer, $a_n$ is a perfect square if and only if $n$ is of the form $\frac{k^2+3k}{2}$ where k is also a positive integer (proof) if n is of the form $\frac{k^2+3k}{2}$ , substituting this into the equation $a_n=9+8n$ gives us, $a_n=9+8(\frac{k^2+3k}{2})$ $a_n= 4k^2+12k+9$ $a_n=(2k+3)^2$ and we are done proving the other direction is easy as we know that $a_n$ must be odd and $(2k+3)^2$ contains all odd square numbers greater than 9 and $a_n>9$ for all n Now let us prove the actual question. We can rearrange the equation given in the question as a quadratic in x. $x^2+x(2y+3)+y^2+y-2n=0$ using the quadratic formula, $x=\frac{-(2y+3)+\sqrt{(2y+3)^2-4(y^2+y-2n)}}{2}$ (note that it cant be $-\sqrt{(2y+3)^2-4(y^2+y-2n)}$ as that would make x negative) if such nonnegative $x$ and $y$ exist, this must mean that (1)$y^2+y\le2n$ (as if $y^2+y-2n>0$, then $\frac{-(2y+3)+\sqrt{(2y+3)^2-4(y^2+y-2n)}}{2}$ will be negative and x will therefore be negative.) also (2) $\sqrt{(2y+3)^2-4(y^2+y-2n)}=t$ where t is a nonnegative integer meaning $(2y+3)^2-4(y^2+y-2n)=8y+9+8n=9+8(y+n)=t^2$ If x and y satisfy these two conditions then we have found pair(x,y) that satisfies the above equation (as$-(2y+3)+\sqrt{(2y+3)^2-4(y^2+y-2n)}$ will always be even anyways) Now let us say that k is a positive integer such that $\frac{(k-1)^2+3(k-1)}{2}<n\le\frac{k^2+3k}{2}$ If we prove that y cant be greater or equal to $\frac{k^2+3k}{2}-\frac{(k-1)^2+3(k-1)}{2}=k+1$ but y can take any value from 1 to k, then we have proved the uniqueness and existence of $(x,y)$ by Claim 1 But this can be done as $\frac{(k-1)^2+3(k-1)}{2}<n$ and so $\frac{(k-1)^2+3(k-1)}{2}+1\le n$ and so $k^2+k \le 2n$ and so we know letting y equal any value from 1 to k will still allow the inequality $y^2+y\le2n$ to be satisfied (as $f(x)=x^2+x$ is an increasing function from $x=-0.5$ onwards)(we have proved the existence of (x,y) as the inequality shows we have satisfied condition 1 and also as we can now always find a y such that $n+y=\frac{k^2+3k}{2}$ (satisfying $9+8(y+n)=t^2$ condition 2 by claim 1) however $(k+1)^2+(k+1)=k^2+3k+2$ and we know from$2n\le\frac{k^2+3k}{2}\times2<k^2+3k+2$ and so we know that y cant be greater than k+1 and we have proved the uniqueness of (x,y)

Notice that $(x+y)^2+3x+y=(x+y)(x+y+1)+2x$. So then we have \[ n = \frac{(x+y)(x+y+1)}{2}+x. \]Notice that $\frac{(x+y)(x+y+1)}{2}$ is just the sum of the numbers $\le x+y$. So for a construction let $k$ be the maximal number such that $\frac{k(k+1)}{2} \le n$. Note that $n-\frac{k(k+1)}{2} \le k$ because if this is not true then $k$ wouldn't be the maximal number satisfying the condition. Then obviously let $x=n-\frac{k(k+1)}{2}$ and then $y=k-x$. EDIT: Also $15^2$ posts!

Apart from uniqueness, this is problem 70 in Chapter 6 of Engel's Problem solving strategies.