Find all pairs $(x, y)$ of integers such that $y^{3}-1=x^{4}+x^{2}$.
If $x=0$, we get a solution $(x, y)=(0,1)$. This solution will turn out to be the only one. If $(x, y)$ is a solution then $(-x, y)$ also is a solution therefore we can assume that $x \geq 1$. We add 1 to both sides and factor: $y^{3}=x^{4}+x^{2}+1=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$. We show that the factors $x^{2}+x+1$ and $x^{2}-x+1$ are co-prime. Assume that a prime $p$ divides both of them. Then $p \mid x^{2}+x+1-\left(x^{2}-x+1\right)=2 x$. Since $x^{2}+x+1$ is always odd, $p \mid x$. But then $p$ does not divide $x^{2}+x+1$, a contradiction. Since $x^{2}+x+1$ and $x^{2}-x+1$ have no prime factors in common and their product is a cube, both of them are cubes by a consequence of the fundamental theorem of arithmetic. Therefore, $x^{2}+x+1=a^{3}$ and $x^{2}-x+1=b^{3}$ for some non-negative integers $a$ and $b$.

As $x \geq 1$ the first equation implies that $a>x^{\frac{2}{3}}$. But since clearly $b<a$ we get

$$
x^{2}-x+1=b^{3} \leq(a-1)^{3}=a^{3}-3 a^{2}+3 a-1 \leq a^{3}-3 a^{2}+3 a \leq a^{3}-2 a^{2}=x^{2}+x+1-2 a^{2}<x^{2}+x+1-2 x^{\frac{4}{3}}
$$

when $2 a^{2} \leq 3 a^{2}-3 a$, i.e. $a^{2} \geq 3 a$ which holds for $a \geq 3$. Clearly $a=2$ is impossible and $a=1$ means $x=0$. We got $x^{2}-x+1<x^{2}+x+1-2 x^{\frac{4}{3}}$ which means $0 \leq 2 x-2 x^{\frac{4}{3}}$. Hence $x=1$, but then 3 would be a cube, a contradiction.

Remark. There are many ways to guess the factorization $x^{4}+x^{2}+1=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$. Writing $x^{4}+x+1=\left(x^{2}+1\right)^{2}-x^{2}$ we immediately obtain it. Another way to see it is to write $p(x)=x^{4}+x^{2}+1$ and notice that $p(0)=1 \cdot 1, p(1)=3 \cdot 1, p(2)=7 \cdot 3$, so there probably is a quadratic factor $q(x)$ for which $q(0)=1, q(1)=3, q(2)=7$ (clearly, there cannot be linear factors). It is easy to see then that $q(x)=x^{2}+x+1$ and the factorization can be completed with the long division. One more way is to write $x^{4}+x^{2}+1=$ $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)$ and compare the coefficients.
It is well-known that if $s$ and $t$ are co-prime integers whose product is a perfect $k$ th power, then $s$ and $t$ both are perfect $k$ th powers. The proof goes like this. Let $a=p_{1}^{\alpha_{1}} \ldots p_{h}^{\alpha_{h}}, b=q_{1}^{\beta_{1}} \ldots q_{\ell}^{\beta_{\ell}}$ and $x=r_{1}^{\gamma_{1}} \ldots r_{m}^{\gamma_{m}}$ be the prime factorizations of $a, b$ and $x$. In the prime factorization of $x^{k}$, every exponent is divisible by $k$, so the same must hold for the factorization of $s t$. But $s$ and $t$ are co-prime, so the exponents of primes in $s$ and $t$ must be divisible by $k$. Therefore $s$ and $t$ are perfect $k$ th powers.