A quartic diophantine

We claim the only solutions are $(0, 1), (0, -1), (-1, 1), (-1, -1)$.

We actually prove a stronger claim: We will prove that the equation $$ y^2 = 1 + x + x^2 + x^3 + x^4 $$ has integer solutions if and only if $x = 0, -1, 3$.

We prove this by showing that the RHS lies strictly between two consecutive perfect squares for $|x| > 1$, unless $x = 3$.

Specifically, we claim that for $|x| > 1, x \neq 3$:

$$ \left(x^2 + \left\lfloor \frac{x}{2} \right\rfloor \right)^2 < 1 + x + x^2 + x^3 + x^4 < \left(x^2 + \left\lfloor \frac{x}{2} \right\rfloor + 1 \right)^2 $$

We split into two cases based on the parity of $x$.

Case 1: $x$ is even

Estimate the RHS: $ x^4 + \frac{x^2}{4} + 1 + x + 2x^2 + x^3 > 1 + x + x^2 + x^3 + x^4 $

This simplifies to: $ 0 < \frac{5x^2}{4} $ which is clearly true for all $x \neq 0$.

Now estimate the lower bound: $ x^4 + \frac{x^2}{4} + x^3 < x^4 + x^3 + x^2 + x + 1. $

This gives: $\frac{3x^2}{4} + x + 1 > 0 $ which is always true since the discriminant is negative.

Case 2: $x$ is odd

Estimate the lower bound: $ x^4 + x^3 – x^2 + \frac{(x-1)^2}{4} < 1 + x + x^2 + x^3 + x^4. $

Simplifying: $ 2x^2 – \frac{(x-1)^2}{4} + x + 1 > 0.$

Expanding and collecting terms: $ 8x^2 – (x^2 – 2x + 1) + 4x + 4 = 7x^2 + 6x + 3 > 0 $ which again is always true (negative discriminant).

Now estimate the upper bound: $ \left(x^2 + \frac{x-1}{2} + 1\right)^2 > 1 + x + x^2 + x^3 + x^4. $

Simplifying shows this reduces to checking: $ |x – 1| > 2 $ which is true unless $x = 3$. In that case:

$ 1 + x + x^2 + x^3 + x^4 = 1 + 3 + 9 + 27 + 81 = 121 = 11^2. $

Thus, $x = 3$ gives a valid solution. In all other cases with $|x| > 1$, the original expression lies strictly between consecutive squares, completing the proof.