Switch the powers

Rewrite the equation as $x^{2} - y^{2} = x^{3} - y^{3}$

Factor it to obtain $(x - y)*(x + y) = (x - y)*(x^{2} + xy + y^{2})$

Given $x - y$ cannot be $0$ , we have: $x + y = x^{2} + xy + y^{2}$

Rearranging all terms in one side, $x^{2} + x*(y - 1) + (y^{2} - y) = 0$

Solve for $x$ .

$x = ((1 - y) +- sqrt(y^{2} - 2y + 1 - 4y^{2} + 4y))/2$

$x = ((1 - y) +- sqrt(y^{2} + 2y + 1 - 4y^{2}))/2$

$x = ((1 - y) +- sqrt((y + 1)^{2} - (2y)^{2}))/2$

$x = ((1 - y) +- sqrt((1 + y)(1 - y))/2$

If $x$ is an integer, then the product $(1 + y)*(1 - y)$ must be greater than or equal $0$ , otherwise we will get a complex number.

Thus, from the inequality we can write: $- 1 <= y <= 1$ . This implies that $y$ is either $-1, 0$ or $1$ .

Case I: $y = - 1$ .

$x - 1 = x^{2} - x + 1$

$x^{2} - 2x + 2 = 0$

$x = (2 +- 2i)/2 = (1 +- i)$ , not valid solutions.

Case II: $y = 0$

$x = x^{2}, x = 0$ or $x = 1$ . Since $y$ is different from $x$ , only the solution $(1, 0)$ is valid.

Case III: $y = 1$

$x + 1 = x^{2} + x + 1$

$x^{2} = 0$ , so $x = 0$ and we have our second solution, $(0, 1)$ .

We will at first ignore the requirement that $x$ and $y$ must be integers, and manipulate this equation algebraically.

$x^2 + y^3 = x^3 + y^2 \:\Rightarrow\: x^2 - y^2 = x^3 - y^3$ $\Rightarrow\: (x - y)(x + y) = (x - y)(x^2 + xy + y^2)$

Because we are given that $x \neq y$ , we know that $x - y \neq 0$ , and so we can divide both sides of the equation by $(x - y)$ , and then we can solve for $x$ in terms of $y$ .

$x + y = x^2 + xy + y^2 \:\Rightarrow\: (1)x^2 + (y - 1)x + (y^2 - y) = 0$

By the quadratic formula,

$x = \frac{1 - y \pm \sqrt{-3y^2 + 2y + 1}}{2}$

Because of the negative leading coefficient, we know that $f(y) = -3y^2 + 2y + 1$ is a parabola opening downwards, and so by finding the vertex of this parabola, we can find all possible nonnegative integer values of $f(y)$ . We find the vertex to be at $y = -\frac{2}{-6} = \frac{1}{3}$ , and so now we only have to check cases for the nonnegative integers less than $f(\frac{1}{3}) = 1\frac{1}{3}$ , which are 1 and 0.

$-3y^2 + 2y + 1 = 1 \:\Rightarrow\: y(-3y + 2) = 0 \:\Rightarrow\: (x, y) = (1, 0),$

$-3y^2 + 2y + 1 = 0 \:\Rightarrow\: (3y + 1)(1 - y) = 0 \:\Rightarrow\: (x, y) = (0, 1)$

So there are $\boxed{2}$ solutions.

First of all, since the equation is symmetrical, if we have some solution $(a,b)$ then $(b,a)$ is also a solution. So if the number of solutions where $x<y$ is $n$ , then the total number of solutions is $2n$ . So therefore let us consider when $x<y$ ie. $y=x+d$ , where $d \in \mathbb{N}$ . Then we can turn the equation into $x^2 + (x+d)^3 = x^3 + (x+d)^2$ . Expanding and cancelling leaves us with: $3x^2 + (3d-2)x + (d^2-d) = 0$ . Applying the quadratic formula we obtain: $x = \dfrac{2-3d \pm \sqrt{9d^2-12d+4-12d^2+12d}}{6}$ $x = \dfrac{2-3d \pm \sqrt{4-3d^2}}{6}$ . Since $x \in \mathbb{R}$ then $4-3d^2 \ge 0$ so the only possible value is $d=1$ . Now we can sub this back in: $x^2+(x+1)^3 = x^3 + (x+1)^2$ , and again expanding and cancelling we get $3x^2 + x = 0$ so $x(3x+1) = 0$ , so the only integral value for x is $0$ , yielding $y = 1$ . As we stated at the start this is only half the solutions, so the final answer is $\fbox {2}$