Integers as solutions! #2

$\begin{aligned} x^4+x^2+2&=y^2+y \\ 4x^4+4x^2+8&=4y^2+4y \\ 4x^4+4x^2+1+8&=4y^2+4y+1 \\ (2x^2+1)^2+8 &= (2y+1)^2 \end{aligned}$

The only square numbers with a difference of $8$ are $1$ and $9$ . Hence the only possible integer solutions are when $(2x^2+1)^2=1$ and $(2y+1)^2=9$ , ie the pairs $(0,-2)$ and $(0,1)$ . So there are just $\boxed2$ solutions in integers.

$x^4+x^2+2=y^2+y$

$\Leftrightarrow y(y+1) = x^4+x^2+2$ $(*)$

It's true that $x^4+x^2 < x^4 + x^2 + 2 < x^4+x^2 + 2 + (4x^2+4)$

$\Leftrightarrow x^2(x^2+1) < y(y+1) < (x^2+2)(x^2+2+1)$

We have this property:

If $x,y,k$ are integers satisfying $x(x+k) < y(y+k) < (x+2)(x+2+k)$ , then $y(y+k)=(x+1)(x+1+k)$

According to the property above, we can conclude that $y(y+1)=(x^2+1)(x^2+2) = x^4+3x^2+2$ $(**)$

Substitute $(**)$ in $(*)$ , we have $x^4+3x^2+2=x^4+x^2+2 \Leftrightarrow x=0$

Substitute $x=0$ in $(**)$ , we have $y(y+1) =2 \Leftrightarrow y^2+y-2=0 \Leftrightarrow y \in \{ -2,1 \}$

Therefore, we have $(x,y) = (0,-2), (0,1)$ . Substitue these results into $(*)$ , these solutions are correct.

Hence there are $\boxed{2}$ different solutions, that is $(x,y) = (0,-2), (0,1)$ .