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!

Solution 1:

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 $.