<s>[INST] Diophantine equation with quartic polynomial 
What are all integral solutions to
$$y^2=x^4+x^3+x^2+x+1$$

The RHS could become
$$(x^2-x+1)(x^2+x+1)+x(x^2+1)$$
or
$$\frac{x^5-1}{x-1}$$
I have no idea how to manipulate the equation into something useful or what the first step should be.
Also, A quartic diophantine equation looks useful, but none of the answers completely solve the question? Thanks!
 [/INST] If $x \gt 0$,
the only solution is
$x=3, y=11$.
Note:
Wolfy was used extensively
getting this answer.
$y^2
=x^4+x^3+x^2+x+1
$
$y^2
=x^4+x^3+x^2+x+1
\gt x^4
\implies
y > x^2
$.
$(x+1)^4
=x^4+4x^4+,,,
$
so
$y^2
\lt (x+1)^4
$
or
$y < (x+1)^2
$.
$(x^2+x/2+3/8)^2
=x^4 + x^3 + x^2 + (3 x)/8 + 9/64
\lt y^2
$
so
$y > x^2+x/2+3/8$.
$(x^2+x/2+1)^2
=x^4 + x^3 + (9 x^2)/4 + x + 1
\gt y^2
$
so
$y < x^2+x/2+1
$.
If $x = 2n$ then
$4n^2+n+3/8
\lt y
\lt 4n^2+n+1
$,
so there can be no such
integer $y$.
If $x = 2n+1$ then
$x^2+x/2+3/8
=4n^2+4n+1+n+1/2+3/8
=4n^2+5n+15/8
$
and
$x^2+x/2+1
=4n^2+4n+1+n+1/2+1
=4n^2+5n+5/2
$
so
$y
= 4n^2+5n+2
$.
But
$y^2
=16 n^4 + 40 n^3 + 41 n^2 + 20 n + 4
$
and
$x^4+x^3+x^2+x+1
=\dfrac{x^5-1}{x-1}
=\dfrac{(2n+1)^5-1}{2n}
=16 n^4 + 40 n^3 + 40 n^2 + 20 n + 5
$
and the difference is
$n^2-1$
so they are never equal
unless $n = 1$
so $x = 3
$.
</s>