Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers

Question

Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers

Answer 1

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 $(11, 3)$.

Answer 2

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 $k \ge 1$.