Diophantine Equation

There's a nice method of solving such equations when the variables are non-negative or positive (could as well be used in other cases, like, e.g., $-x^2-y^2=z^2$ for $x,y,z\in\mathbb R$ , but such cases are a bit more rare).

Write the equation as $\displaystyle -y-xy=x^2-x+y^2$ .

Since $x^2\ge x$ , we have that the RHS is non-negative and the LHS is non-positive, hence $\displaystyle\begin{cases}-y-xy=0\\x^2-x+y^2=0\end{cases}\implies \begin{cases}-y(1+x)=0\\x\ge 0\end{cases}\implies y=0.$

Therefore, $y=0$ , which means $x_1=0$ and $x_2=1$ .

So the solutions are $(0,0)$ and $(0,1)$ (after checking them, they work), which means our answer is $0+0+0+1=\boxed{1}$ .

There are other ways of solving this (I have some other solutions that solve this), but this one uses a nice technique I wanted to show you.

In this solution ,the the approach is to write the equation in simple form.

$\displaystyle x-y=x^2+xy+y^2$

$\Rightarrow\displaystyle (x-y)+xy=(x^2+xy+y^2) +xy$

$\Rightarrow\displaystyle (x-y)+xy=(x+y)^2$

$\Rightarrow\displaystyle -1+(x-y)+xy=(x+y)^2 -1$

$\Rightarrow\displaystyle (x-1)(y+1)=(x+y)^2 -1$

Now let $x-1=a$ and $y+1=b$ .

We find that $x+y=a+b$ .

So the equation can be written as

$\displaystyle ab=(a+b)^2 -1$

$\Rightarrow\displaystyle ab=a^2+b^2+2ab -1$

$\Rightarrow\displaystyle 1=a^2+b^2+ab$

Since $a^2,b^2$ , and $ab \geq-1$

The only solution of above are $(a,b) \in (0,1),(1,0),(-1,1),(1,-1)$

So $(x,y)\in(2,-1),(0,0),(1,0),(2,-2)$ but since $x,y\geq0.$

The solutions are $(x,y)\in(0,0),(1,0)$ .

So the answer is $\boxed1$

put x = y + k. we get a quadratic in y. find its discriminant. it equals 3k(4-k) this must be a perfect square. hence k =0,1,3,4. plug in and look for non negative solutions.