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
 A: 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 $(11, 3)$.
A: I will look  at
$x(x+1) = y(y+1)(y^2+k)$
for integral $k \ge 1$.
This becomes the original question
when $k = 2$.
I will show that
there are no solutions in
positive integral $x$ and $y$
for $y \ge k+2$.
Note  that
$(x, y)
=(k^2-k, k-1)$
and
$(k^2+3k+1, k+1)
$
are solutions to this,
and there is no solution
with $y = k$.
These correspond to
the solutions
$(x, y) = (2, 1)$ and
$(11, 3)$
to the original equation.
This is essentially
my previous solution for
$k=2$
with slightly more complicated algebra.
$x(x+1) = y(y+1)(y^2+k)
=y(y^3+y^2+ky+k)
=y^4+y^3+ky^2+ky
$
Multiplying by 4,
$(2x+1)^2-1 
=4y^4+4y^3+4ky^2+4ky 
$
or
$(2x+1)^2
=4y^4+4y^3+4ky^2+4ky+1 
$
My goal is to show algebraically 
that this polynomial in $y$
is between two consecutive squares
for large enough $y$.
$(2y^2+y)^2
=4y^4+4y^3+y^2
$.
$\begin{align} 
(2y^2+y+k)^2 
&=4y^4+4y^3+y^2 +2k(2y^2+y)+k^2 \\ 
&=4y^4+4y^3+(4k+1)y^2+2ky+k^2 \\ 
\end{align} 
$. 
$\begin{align}
(2y^2+y+k-1)^2 
&=4y^4+4y^3+y^2
+2(k-1)(2y^2+y)+(k-1)^2 \\
&=4y^4+4y^3+(4k-3)y^2+(2k-2)y+(k-1)^2 \\
\end{align}
$.
For
$(2x+1)^2$
to be between these consecutive squares,
we need
$(4k-3)y^2+(2k-2)y+(k-1)^2
<4ky^2+4ky+1
<(4k+1)y^2+2ky+k^2
$.
The first inequality is
$0
<3y^2+(2k+2)y-(k-1)^2+1
$
or
$y(3y+2k+2)
>k(k-2)
$
and this is certainly true for
$y \ge k$.
For the second inequality to be true,
we need
$4ky^2+4ky+1 
<(4k+1)y^2+2ky+k^2 
$
or
$y^2-2ky+k^2-1
> 0
$
or
$(y-k)^2-1
> 0$.
This is true for
$y \ge k+2$,
so the equation has no solution for $y \ge k+2$.
If
$y=k-1$,
the right side is
$k(k-1)((k-1)^2+k)
=(k^2-k)(k^2-k+1)
$,
so $x=k^2-k$, $y=k-1$
is a solution.
Similarly, if 
$y=k+1$, 
the right side is 
$(k+1)(k+2)((k+1)^2+k) 
=(k^2+3k+2)(k^2+3k+1) 
$, 
so $x=k^2+3k+1$, $y=k+1$ 
is  a solution.
If $y=k$,
the equation is
$x(x+1)
= k(k+1)(k^2+k)
= (k^2+k)^2
$,
or
$(2x+1)^2-1
= (2k^2+2k)^2 
$, 
which has no solutions for
$k \ge 1$.
