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

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 $y$.
