Quick! Bound it!

Note the following

$\begin{aligned} y^2(x^2+y^2-2xy-x-y)&=(x+y)^2(x-y)\\ y^2(x-y)^2-y^2(x+y)&= (x+y)(x^2-y^2)\\ y^2(x-y)^2&= (x+y)(x^2-y^2)+(x+y)y^2\\ y^2(x-y)^2 &= x^2 (x+y) \end{aligned}$

Note that because the LHS is a perfect square, the RHS must also be a perfect square. Thus, $x+y$ is a perfect square. Let $x+y=k^2$ or $k^2-x=y$ . (We will first check when $x \geq y$ )

$\begin{aligned} y^2(x-y)^2&=x^2(x+y)\\ y(x-y)&=kx\\ (k^2-x)(2x-k^2)&=kx\\ 2x^2 - (3k-1)kx + k^4 &= 0 \quad \quad \quad (1) \end{aligned}$

If $x \leq y$ , then we have

$\begin{aligned} y(y-x) &= kx\\ (k^2-x)(k^2-2x) &= kx\\ 2x^2 - (3k+1)kx + k^4 &= 0 \quad \quad \quad (2) \end{aligned}$

Let $\Delta_1$ be the discriminant of $(1)$ and $\Delta_2$ be the discriminant of $(2)$ . We know that both these discriminants have to be perfect squares for us to have integer solutions.

$\begin{aligned} \Delta_1 &= k^2(3k-1)^2 - 8k^4\\ &= k^4-6k^3+k^2\\ &= k^2((k-3)^2-8) \end{aligned}$

The only values of $k$ which satisfy the discriminant being a perfect square are when $k=0, 6$ .

$\begin{aligned} \Delta_2 &= k^2 (3k+1)^2 - 8k^4\\ &= k^4+6k^3+k^2\\ &=k^2((k+3)^2-8) \end{aligned}$

The only values of $k$ which satisfy the discriminant being a perfect square are when $k=-6,0$ .

Thus, $k^2=0, 36$ , which implies $x+y=0, 36$ .

If $x+y=0$ , $x=-y$ . Substituting, we have

$\begin{aligned} x^2 (2x)^2 &= 0\\ 4x^4&=0\\ x&=0, y=0 \end{aligned}$

Thus, $(0,0)$ is a solution in $(x, y)$ . If $x+y=36$ , $y=36-x$ . If $x$ is greater than or equal to 18,

$\begin{aligned} (36-x)^2 (2x-36)^2 &= 36x^2\\ -2x^2+108x-1296 &= 6x\\ x^2-51x+648 &= 0\\ (x-24)(x-27) &= 0\\ x &= 24,27 \end{aligned}$

Thus, the solutions $(x, y)$ when $x$ is greater than or equal to 18 are $(24, 12)$ and $(27, 9)$ . If $x$ is less than or equal to 18,

$\begin{aligned} (36-x)^2 (2x-36)^2 &= 36x^2\\ 2x^2-108x+1296 &= 6x\\ x^2-57x+648=0 \end{aligned}$

This quadratic is unfactorable for integers. Thus, the only solutions or $(x, y)$ are $(0, 0)$ , $(24, 12)$ and $(27, 9)$ .

First, if $x=0$ then, $y^2 ( y^2 - y) = - y^3$ , so $y=0$ if $y=0$ then, $0= x^3$ , so $x=0$
And (0,0) satisfies the above expression

let $x = ga$ and $y = gb$ ( $GCD(a,b) = 1 ~~and ~~let ~~gab ≠0~~and ~b>0 )$

then, $g^3 b^2 ( g(a-b)^2 - (a+b) ) = g^3 (a+b)^2 (a-b)$ , so we get $b^2 ( g(a-b)^2 - (a+b) ) = (a+b)^2 (a-b)$

Because $b^2$ divides LHS , $b^2$ divides RHS and $GCD(a,b)= 1$ , $gcd ( b^2 , (a+b)^2 ) = gcd( b^2 , (a-b) ) = 1$

so $b^2 = 1$ and $b=1$

By " $b=1$ "

$g(a-1)^2 - (a+1) = (a+1)^2 (a-1)$

$g(a-1)^2 = (a+1) a^2$

therefore $(a= 2 ~~and~~ g=12 ~~~~or~~~a=3 ~~and ~~g=9~~) , (x,y)= (24,12) , ~(27, 9)~$
$( ∵ gcd( (a-1)^2 , a^2 ) = 1 so (a-1)^2 | a+1 , ~ so ~ a=2 ~or ~ a=3 )$

By above, $(x,y ) = (0,0)~,~(24,12)~,~(27,9) ~$

As already given by Sharky Kesa:

$\begin{aligned} y^2(x^2+y^2-2xy-x-y)&=(x+y)^2(x-y)\\y^2(x-y)^2-y^2(x+y)&= (x+y)(x^2-y^2)\\y^2(x-y)^2&= (x+y)(x^2-y^2)+(x+y)y^2\\y^2(x-y)^2 &= x^2 (x+y)\end{aligned}$

We see that $x+y$ must be square.

Substitute $\boxed{\begin{aligned}a^2=x+y\\b=x-y\\x=\frac{a^2+b}2\\y=\frac{a^2-b}2\end{aligned}}$ to get $\begin{aligned}\dfrac{(a^2-b)^2}{4}b^2&=\dfrac{(a^2+b)^2}4a^2 \\ (a^2-b)b&=\pm(a^2+b)a \\ a^2b\mp ba&=b^2\pm a^3 \\ a\mp1&=\dfrac{b^2\pm a^3}{ab} \end{aligned}$

Case $a=0$ or $b=0$ : if $a=0$ , then $b=0$ and inversely. And then $x=y=0$ . It is a solution.

Case $a,b,x,y\neq0$ :

We see that $b$ divides $a^3$ and $a$ divides $b^2$ , which means that each primes dividing $a$ also divides $b$ and inversely: $\displaystyle a=\prod p_i^{m_i},~b=\prod p_i^{n_i}\text{ with }m_i,n_i\geq1$

Let consider a prime divisor $p$ so that $a=p^m·a'$ and $b=p^n·b'$ and $(p,a')=(p,b')=1$ . As $p$ doesn't divide the LHS, it doesn't divide the RHS: $\dfrac{p^{2n}·b'^2\pm p^{3m}·a'^3}{p^{m+n}·a'b'} \text{ is an integer not divisible by }p$

If $n<m$ , then $p^{m+n}$ divides $p^{3m}·a'^3$ but not $p^{2n}·b'^2$ , that cannot be an integer.

If $2m<n$ , then $p^{m+n}$ divides $p^{2n}·b'^2$ but not $p^{3m}·a'^3$ , that cannot be an integer.

If $m<n<2m$ , then $p^{2n}·b'^2$ and $p^{3m}·a'^3$ both have at least a factor $p$ after the division: the RHS would then be divisible by $p$ .

Therefore, $n=m$ or $n=2m$ . We can write $\begin{aligned}\displaystyle a=\prod q_i^{m_i}·\prod r_j^{m_j},~b=\prod q_i^{m_i}·\prod r_j^{2m_j} \\ \Longleftrightarrow a=e·f\text{ and }b=e·f^2\text{ for some integers }e,f\end{aligned}$

Put it again into $(a^2-b)b=\pm(a^2+b)a$ to get $\begin{aligned}(e^2·f^2-e·f^2)e·f^2&=\pm(e^2·f^2+e·f^2)e·f \\ (e-1)·f&=\pm(e+1) \end{aligned}$ $f=\pm\dfrac{e+1}{e-1}=\pm\left(1+\dfrac{2}{e-1}\right)$ .

$e-1$ divides 2 and thus can only be $e-1=\pm1$ or $e-1=\pm2$ , which gives $(e,f)=(2,3),(0,-1),(3,2),(-1,0)$ .

Case $e=2,f=3$ : $a=6,b=18,x=27,y=9$

Case $e=3,f=2$ : $a=6,b=12,x=24,y=12$

There are three possibilites : $\boxed{(0,0),~(27,9),~(24,12)}$

We begin by simplifying $y^2(x^2+y^2-2xy-x-y)=(x+y)^2(x-y)=(x^2-y^2)(x+y)$ $y^2(x-y)^2-y^2(x+y)=x^2(x+y)-y^2(x+y)$ $y^2(x-y)^2=x^2(x+y)$ At this point we break the problem up into $2$ cases: $x\ge y$ and $x<y$ . We tackle $x\ge y$ first by taking both sides modulus $y$ , which gives $0\equiv x^2(x+y)\equiv y^3\pmod{y}$ Hence, $x=ny$ , for some integer $n$ . Plugging this into the equation gives $y^2(y(n-1))^2=y^4(n-1)^2=n^2y^2(y(n+1))=n^2y^3(n+1)$ $y=\frac{(n+1)n^2}{(n-1)^2}$ This means that $(n-1)^2\mid ((n+1)n^2)$ . Since these factors differ by $1$ or $2$ , these factors should be relatively small (otherwise they cannot divide each other). Checking the possible values reveals that $n\in\{-1,0,2,3\}$ which gives $y\in\{0,0,12,9\}$ , respectively. Since $x=ny$ , this gives the solutions $(x,y)\in\{(0,0),(24,12),(27,9)\}$ .

Now, if $x<y$ , then taking both sides modulus $x$ gives $y^2(x-y)^2\equiv y^2(-y)^2\equiv y^4\equiv 0\pmod{x}$ so $y=nx$ , for some integer $n$ . Plugging this into the equations gives $n^2x^2(x(1-n))^2=x^4n^2(1-n)^2=x^2(x(1+n))=x^3(1+n)$ $x=\frac{1+n}{n^2(1-n)^2}$ This means that $n^2(1-n)^2)\mid (1+n)$ . Since these factors once again differ by $1$ or $2$ , we find that the only possible value of $n$ is $n=-1$ , which gives $x=0$ , which was already determined in the previous case.

Hence, the only solutions are $\boxed{(x,y)\in\{(0,0),(24,12),(27,9)\}}$ , which means our desired answer is $\boxed{3}$ .