Solve in Z. $y^2+y=x^4+x^3+x^2+x$ Solve in Z. $$y^2+y=x^4+x^3+x^2+x$$
In my attempt to solve this, I used the fact that the left side is even.I write the eqyation in the following form: $$(x+1)(x^2+1)=0 \pmod 2$$. This clearly means x must be an odd number, as a result $$4|y(y+1)$$. This is as far as I could go.

Question from Jalil Hajimir


Hint: Bound between 2 squares
For all but finitely many values of $x$, we can bound $(2y+1)^2$ between 2 consecutive perfect squares, hence it is never a perfect square.

 $(2x^2 + x)^ 2 < (2y+1) ^2 = 4y^2 + 4y + 1 = 4x^4 + 4x^3 + 4x^2 + 4x + 1  < 4x^4 + 4x^3 + 5x^2 + 2x + 1 = (2x^2 + x + 1)^2 $

Hence, we only need to check those values when the inequalities fail: 
RHS inequality is false when: $x \in [0, 2 ] $.
$x = 0$ yield $ y = -1, 0$.
$x = 1$ yields no integer solution for $y$.
$x = 2$ yield $ y = -6, 5$.   
LHS inequality is false when $x \in [-1, -\frac{1}{3} ]$.
$x = -1$ yields $y = -1, 0$.   
