Exercise 1.2.13 (First cases of Little Fermat's Theorem)

Proof. Let $n\in \mathbb{Z}$.

(a)

If $n$ is odd, then $2\mid n-1$, and if $n$ is even, $2\mid n$. In both cases, $2\mid n(n-1)=n^2-n$.

(b)

We proved in Exercise 6 that the product of three consecutive integers is divisible par $6$. Therefore $$6\mid (n-1)n(n+1)=n^3-n.$$

(c)

We first show that $5\mid n^5-n$.

First proof: One of the five consecutive integers $n-2,n-1,n,n+1,n+2$ is divisible by $5$. Therefore

$$5\mid (n-2)(n-1)n(n+1)(n+2).$$

Therefore $5\mid n(n^2-1)(n^2-4)=n(n^2-1)(n^2+1-5)=n(n^2-1)(n^2+1)-5n(n^2-1)$. This shows that $5\mid n(n^2-1)(n^2+1)=n^5-n$.

Second proof : by induction. First $5\mid 0^5-0=0$. Now assume that $5\mid n^5-n$, so that $n^5-n=5k,k\in \mathbb{Z}$. Then, using the binomial formula,

$$\begin{array}{llll}\hfill {(n+1)}^5-(n+1)& =(n^5+5n^4+10n^3+10n^2+5n+1)-(n+1)& \hfill & \\ \hfill & =(n^5-n)+5(n^4+2n^3+2n^2+n)& \hfill & \\ \hfill & =5(k+n^4+2n^3+2n^2+n).& \hfill & \end{array}$$

Therefore $5\mid {(n+1)}^5-(n+1)$, and the induction is done. We can conclude, for all $n\in \mathbb{N}$,

$$5\mid n^5-n.$$

Moreover, if $n\in {\mathbb{Z}}^{\text{−}}$, then $m=-n\in N$, thus $5\mid m^5-m=-(n^5-n)$, so $5\mid n^5-n$:

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{Z},5\mid n^5-n.$$

Conclusion: Since $6\mid (n-1)n(n+1)$ by part (b) and

$$(n-1)n(n+1)\mid n^5-n=n(n^2-1)(n^2+1)=[(n-1)n(n+1)](n^2+1),$$

we obtain $6\mid n^5-n$.

Since $6$ and $5$ are relatively prime, for all $n\in \mathbb{N}$,

$$30\mid n^5-n.$$

Note: $2\mid n^2-n,3\mid n^3-n,5\mid n^5-n$ are particular cases of Little Fermat’s Theorem.