Show that $x(x+1) = y^4+y^3+ay^2+by+c$ has a finite number of positive integral solutions.

Question

More precisely, If $a$, $b$, and $c$ are integers, show that there are only a finite number of positive integers $x$ and $y$ such that $x(x+1) = y^4+y^3+ay^2+by+c$.

I have a solution, which I will show in two days if no better one is found.

Answer 1

Here is my answer. Interestingly, it uses no divisibility properties.

We are looking at $x(x+1) = y^4+y^3+ay^2+by+c$.

I will show that, if $a$, $b$, and $c$ are integers, there are at finite number of solutions in positive integral $x$ and $y$. This will be done by finding bounds for $y$ in terms of $a$, $b$, and $c$.

Multiplying by 4, $(2x+1)^2-1 =4y^4+4y^3+4ay^2+4by+4c $

or $(2x+1)^2 =4y^4+4y^3+4ay^2+4by+d $, where $d = 4c+1$.

My goal is to show algebraically that this polynomial in $y$ is between two consecutive integer squares for large enough $y$.

First,

$(2y^2+y)^2 =4y^4+4y^3+y^2 $.

Next,

$\begin{align} (2y^2+y+a)^2 &=4y^4+4y^3+y^2 +2a(2y^2+y)+a^2 \\ &=4y^4+4y^3+(4a+1)y^2+2ay+a^2 \\ \end{align} $.

Finally,

$\begin{align} (2y^2+y+a-1)^2 &=4y^4+4y^3+y^2 +2(a-1)(2y^2+y)+(a-1)^2 \\ &=4y^4+4y^3+(4a-3)y^2+(2a-2)y+(a-1)^2 \\ \end{align} $.

Since $(2x+1)^2 =4y^4+4y^3+4ay^2+4by+d $, for $(2x+1)^2$ to be between these consecutive squares, we need

$(4a-3)y^2+(2a-2)y+(a-1)^2 <4ay^2+4by+d <(4a+1)y^2+2ay+a^2 $.

I will now show that both inequalities are true for large enough $y$.

The first inequality is the same as $0 <3y^2+(4b-2a+2)y+d-(a-1)^2 $

or $0 <9y^2+6(2b-a+1)y+3(d-(a-1)^2) $

or $0 < (3y-(2b-a+1))^2-(2b-a+1)^2 +3(d-(a-1)^2) $

or $ (3y-(2b-a+1))^2 >(2b-a+1)^2 -3(d-(a-1)^2) $

and this is certainly true for $y$ large enough.

For the second inequality to be true, we need $4ay^2+4by+d <(4a+1)y^2+2ay+a^2 $

or $y^2+(2a-4b)y+a^2-d > 0 $

or $(y+(a-2b))^2-(a-2b)^2+a^2-d > 0 $

or $(y+(a-2b))^2 >(a-2b)^2-a^2+d =a^2-4ab+4b^2-a^2+d =4b^2-4ab+d $.

This is true for $y$ large enough, so the equation has no solutions for large enough $y$.