# 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

• You should try to improve your question by putting on it what you have tried so far. People here will be more interested in help you, if they see that you are interested in learn, instead of just getting the solution of this problem. – Tomás Sep 13 '13 at 13:40
• Show some effort towards your own question. Some rather boring solutions are $\;(0,0)\;,\;(0,-1)\;,\;(-1,0)\;,\;(2,1)\;,\ldots$ – DonAntonio Sep 13 '13 at 13:43

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 $(11, 3)$.

• What happened to $(x,y)=(2,1)$ ? – Dietrich Burde Sep 13 '13 at 14:02
• I'm fixing that up right now. – marty cohen Sep 13 '13 at 14:14
• It's not that much harder to extend your argument to the negative numbers. Your inequality will be true for large enough $|y|$ – Calvin Lin Sep 13 '13 at 14:54
• That is left as an exercise for the reader. Also see my generalization. – marty cohen Sep 13 '13 at 15:46

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 $k \ge 1$.