Who's up to the challenge? 24

The given equation can be written as $x^{2} + y = 5x - 5y^{2} \Longrightarrow x^{2} - 5x + y + 5y^{2} = 0$ .

As this is a quadratic in $x$ , we see that

$x = \dfrac{5 \pm \sqrt{(-5)^{2} - 4(y + 5y^{2})}}{2} = \dfrac{5 \pm \sqrt{-20y^{2} - 4y + 25}}{2}$ .

Now in order for $x$ to be an integer we will require that for some integer $n \ge 1$

$-20y^{2} - 4y + 25 = n^{2} \Longrightarrow -20(y^{2} + \frac{1}{5} + \frac{1}{100}) + \frac{1}{5} + 25 = n^{2}$

$\Longrightarrow -20(y + \frac{1}{10})^{2} + \frac{126}{5} = n^{2}$ .

(Note that we can't have $n = 0$ as this would make $x = \frac{5}{2}$ , i.e., non-integral.) Now as $n^{2} \ge 1$ we will then require that

$-20(y + \frac{1}{10})^{2} \ge 1 - \frac{126}{5} \Longrightarrow (y + \frac{1}{10})^{2} \le \frac{121}{100} \Longrightarrow |y + \frac{1}{10}| \le \frac{11}{10} \Longrightarrow -\frac{6}{5} \le y \le 1$ .

Since $y$ must also be integral we see that the only possible values for $y$ are $-1, 0$ and $1$ . Plugging these values into our quadratic solution for $x$ yields the potential solution pairs

$(1,-1), (4,-1), (0,0), (5,0), (2,1)$ and $(3,1)$ .

Checking each of these back in the original equation, we see that $(1,-1)$ and $(0,0)$ yield indeterminate values for the expression $\frac{x^{2} + y}{x - y^{2}}$ , leaving us with a final total of $\boxed{4}$ valid integral solution pairs $(x,y)$ .

$\large \dfrac { { x }^{ 2 }+y }{ x-{ y }^{ 2 } } =5,\ \ \implies\ \ \ 5y^2+y+x^2 - 5x=0.\\ Solving\ as\ a\ quadratic\ in\ x, \ \ x= \dfrac{5 \pm\ \sqrt{25 - 20y^2 - 4y}}{2}.\\ We\ can\ clearly\ see\ that\ for\ even \ a\ real \ x,\ integer\ y\ \ has\ to\ be -1\ \leq\ y\ \leq\ 1.\\ y= - 1, \ \ x=\dfrac{5 \pm \sqrt{9}}2=4\ \ OR\ \ 1, \ but\ 1\ is\ not\ valid\ so \ \ \ we\ have\ \ (4, - 1)\ as\ solution..............\{1\}\\ y=\ \ 0,\ \ \ x=\dfrac{5 \pm \sqrt{25}}2=5\ \ OR\ \ 0, \ but\ 0\ is\ not\ valid\ so \ \ \ we\ have\ \ (5, \ \ 0)\ as\ solution.............\{2\}\\ y= \ \ 1,\ \ \ x=\dfrac{5 \pm \sqrt{1}}2=3\ \ OR\ \ 2, \ \ both\ are\ valid\ so \ \ \ we\ have\ \ (2,1)\ and\ \ (3,1)\ \ as\ solutions.....\{4\} \\ \Large\ \ \color{#D61F06}{4}\ \ \ solutions.$