183 Find all integer solutions to the indeterminate equation $y^{2}=1+x+x^{2}+x^{3}+x^{4}$.
The original indeterminate equation can be rearranged as
$$
\left(x^{2}+\frac{x}{2}+\frac{\sqrt{5}-1}{4}\right)^{2}=y^{2}-\frac{5-2 \sqrt{5}}{4}\left(x+\frac{3+\sqrt{5}}{2}\right)^{2}
$$

and
$$
\left(x^{2}+\frac{x}{2}+1\right)^{2}=y^{2}+\frac{5}{4} x^{2} .
$$

Since $x^{2}+\frac{x}{2}+\frac{\sqrt{5}-1}{4}$ and $x^{2}+\frac{x}{2}+1$ are both positive, from (1) and (2) we get
$$
x^{2}+\frac{x}{2}+\frac{\sqrt{5}-1}{4} \leqslant|y| \leqslant x^{2}+\frac{x}{2}+1 .
$$

We can set $|y|=x^{2}+\frac{x+a}{2}, 0<a \leqslant 2$. Since $x, y$ are integers, when $2 \mid x$, $a=2$, so substituting $|y|=x^{2}+\frac{x}{2}+1$ into (2) gives $x=0$, and thus $y= \pm 1$; when $2 \nmid x$, $a=1$. Transforming (2) into
$$
y^{2}=\left(x^{2}+\frac{x+1}{2}\right)^{2}-\frac{(x-3)(x+1)}{4} \text {. }
$$

Substituting $|y|=x^{2}+\frac{x+1}{2}$ into the above equation, we get $x=3$ or $x=-1$, and thus the solutions are $x=-1, y= \pm 1 ; x=3, y= \pm 11$. Therefore, all solutions to the equation are: $x=-1, y= \pm 1 ; x=0, y= \pm 1$; $x=3, y= \pm 11$.