Fascinating Diophantines

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

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

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

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

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

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

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

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 $(x,y)\in(1,0),(2,1),(0,0),(2,2)$ .

So the answer is $(1+0+2+1+0+0+2+2)+4=\boxed{12}$ .

The expression $x+y=x^2-xy+y^2$ can be written as $x^2+y^2-xy-x-y=0$ .Now multiplied both side of the new equation by $2$ and rewrite the expression as follows $x^2-2xy+y^2 +x^2-2x+1 +y^2-2y+1=2$

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

$\Rightarrow$ we get three cases.

Case $1:$ $\ { x-y }=0,x=1,-1 ,y=1,-1$ here accepted solution are $(0,0),(2,2)$

Case $2:$ $x-y=1,-1 ,x-1=0, y-1=1,-1$ here accepted solution are $(1,0),(1,2)$

Case $3:$ $x-y=1,-1, x-1=1,-1, y-1=0$ here accepted solution are $(0,1),(2,1)$

but as stated in the question $(1,0),(1,2)$ $=$ $(0,1),(2,1)$ .Hence answer would be $12$ .

Rearranging gives $x^2 + x(-y - 1) + (y^2- y) = 0$ . Since we want to find integer roots, then quadratic discriminant (in $x$ ) must be a square of an integer (and non-negative):

$(-y-1)^2 - 4(1)(y^2 - y) \geq 0 \quad \Leftrightarrow \quad 3y^2 - 6y - 1 \leq 0 \quad \Leftrightarrow \quad \dfrac{3 - 2\sqrt3}3 \leq y \leq \dfrac{3 + 2\sqrt3}3,$

so the only possible values of $y$ are 0, 1 and 2.

Trial and error shows that $(x,y) = (0,0), (1,0), (0,1) , (2,1) , (1,2)$ .