Solve for $x \in \mathbb{Z}$, the equation $x^{4}-2=x^{3}+x^{2}+x$
If $x=1$, we have no solution, so we can assume $x \neq 1$

$$
\begin{gathered}
x^{4}-2=x^{3}+x^{2}+x \Longleftrightarrow x^{4}-1=x^{3}+x^{2}+x+1 \Longleftrightarrow x^{4}-1=\frac{x^{4}-1}{x-1} \\
\Longleftrightarrow x-1=1 \text { or } x^{4}-1=0 \Longleftrightarrow x=2 \text { or } x=-1
\end{gathered}
$$

We verify that $x=2$ and $x=-1$ are indeed solutions, so these are the only ones.

## Practical Tips

Definition 83. Let $p$ be a prime number. The "p-adic valuation" of an integer $n$, denoted $v_{p}(n)$, is the highest power of $p$ that divides $n$.

Example 84. $v_{2}(42)=1 ; v_{5}(42)=0 ; v_{3}(54)=3$

Lemma 85. For all integers $a$ and $b$, and $p$ prime, $v_{p}(a b)=v_{p}(a)+v_{p}(b)$

Proof. By the prime factorization of $a$ and $b$

Theorem 86. If an integer $a$ can be written in the form $a=x^{n}$, then for any prime number $p$ dividing $n$, there exists an integer $k$ such that $v_{p}(a)=k n$

Proof. $v_{p}(a)=v_{p}\left(x^{n}\right)=n \times v_{p}(x)$. It is therefore sufficient to choose $k=v_{p}(x)$