<s>[INST] Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers
 [/INST] Here's a solution for positive $x$ and $y$.
I will show that
the only solutions 
for positive $x$ and $y$ are
$(x, y) = (2, 1)$ and $(11, 3)$.
$x(x+1) = y(y+1)(y^2+2)
=y(y^3+y^2+2y+2)
=y^4+y^3+2y^2+2y
$
Multiplying by 4,
$(2x+1)^2-1 
=4y^4+4y^3+8y^2+8y 
$
or
$(2x+1)^2
=4y^4+4y^3+8y^2+8y+1 
$
My goal is to show algebraically 
that this polynomial in $y$
is between two consecutive squares
for large enough $y$,
and then examine the remaining cases.
$(2y^2+y)^2
=4y^4+4y^3+y^2
$.
$\begin{align}
(2y^2+y+1)^2 
&=4y^4+4y^3+y^2
+2(2y^2+y)+1 \\
&=4y^4+4y^3+y^2 +4y^2+2y+1 \\
&=4y^4+4y^3+5y^2+2y+1 \\
\end{align}
$.
$\begin{align}
(2y^2+y+2)^2 
&=4y^4+4y^3+y^2 +4(2y^2+y)+4 \\
&=4y^4+4y^3+y^2 +8y^2+4y+4 \\
&=4y^4+4y^3+9y^2+4y+4 \\
\end{align}
$.
For
$(2x+1)^2$
to be between these consecutive squares,
we need
$5y^2+2y+1
<8y^2+8y+1 
<9y^2+4y+4 
$.
The first inequality is true
for $y \ge 1$.
For the second inequality to be true,
we need
$8y^2+8y+1 
<9y^2+4y+4 
$
or
$y^2-4y+3
> 0
$
or
$(y-2)^2-1
> 0$.
This is true for
$y \ge 4$,
so the equation has no solution for $y \ge 4$.
If $y = 3$,
the equation is
$x(x+1) = 3(4)(11)$
and this is true for
$x=11$
(surprise!).
If $y = 2$,
the equation is
$x(x+1) = 2(3)(4)=24$
which has no solution.
If $y = 1$,
the equation is
$x(x+1) = 1(2)(3)$
and this is true for
$x=2$.
Therefore
the only solutions 
for positive $x$ and $y$ are
$(x, y) = (2, 1)$ and $(11, 3)$.
</s>