Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$

Question

Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.

Answer 1

Hint

It suffices to consider the cases $n=1,2,3$ or $4$.

Answer 2

I would approach this with a proof by cases. There are $5$ options for $n\pmod{5}$:

Case $n\equiv 0\pmod 5$: Not possible by assumption.

Case $n\equiv 1 \pmod 5$: In this case, note that $n^4\equiv 1^4 \equiv 1 \pmod 5$

Case $n\equiv 2 \pmod 5$: (keep going--it's similar...)

Case ...

EDIT: this does assume that you meant "divisible," not "a divisor."

Answer 3

$$n^4-1=(n-1)(n+1)(n^2+1)$$

Since $\;n\neq0\pmod 5\;$, if it is $\;\pm1\pmod 5\;$ the above is $\;0\pmod 5\;$, and if it is $\;n=\pm 2\pmod 5\;$, then $\;n^2=(\pm2)^2=4\implies n^2+1=0\pmod 5\;$.

In any case, $\;n^4-1=0\pmod 5\;$

Answer 4

Fermat's little theorem states that if $p$ is prime, then for all $n$ with $\gcd(n,p)=1$, we have $$n^{p-1} \equiv 1 \mod p.$$ In your case we have $p=5$.

One proof: Because $\gcd(n,p)=1$ there is a number $m$ such that $nm \equiv 1 \mod p$. This means the map $f(x)=nx$ is a bijection on the set $\{1,2,\dots, p-1\}$ to itself since there is an inverse $f^{-1}(x)=mx$. Therefore,

$$1n \cdot 2 n\cdots (p-1)n = 1 \cdot 2 \cdots (p-1) \mod p$$

because the same set of numbers mod $p$ is being multiplied on each side. Rewrite

$$n^{p-1} (p-1)! \equiv (p-1)! \mod p$$

and cancel the $(p-1)!$ terms, which we can do because every number in $\{1,2,\dots, p-1\}$ is relatively prime to $p$ and has an inverse.

Answer 5

$$n \equiv \pm 1 \space\mathrm{or} \pm 2 \pmod 5$$

Hence

$$n^4 \equiv (\pm 1)^4 \space\mathrm{or} \space(\pm 2)^4 \pmod 5 \\ \implies n^4 \equiv 1 \space\mathrm{or} \space 16 \pmod 5 \\ \therefore n^4 \equiv 1 \pmod 5$$

Answer 6

Like Timbuc,

$$n^4-1=(n-1)(n+1)(n^2+1)=(n-1)(n+1)(n^2-4+5)$$

$$=\underbrace{(n-2)(n-1)(n+1)(n+2)}+5(n-1)(n+1)$$

As $n$ must divide exactly one of any five consecutive integers and $5\nmid n,(5,n)=1,$

$n$ must divide one of the multiplicand underbrace

Answer 7

Here's a group theoretic perspective. The set of positive integers smaller than and relatively prime to $5$ form a group under multiplication mod $5$. If $n$ is not divisible by $5$ then $n$ is not congruent to $0$ mod $5$. Thus working mod $5$ we have $n$ is congruent to one of $\{1,2,3,4\}$ which is precisely the group mentioned above. The order of this group is $4$, and as a corollary to lagrange's theorem any member of the group raised to the $4^{th}$ power must be the identity element, namely $1$. Hence $n^4 \equiv 1$ mod $5$.

Answer 8

Here's one particularly pretty way of proving this statement. It's nice because it uses combinatorics to prove an essentially number theoretic result. We start with the problem:

Suppose you want to make a necklace with $5$ beads, and you can paint each one of them with one of $n$ available colors. What's the number of different necklace you can have? (Necklace that can be obtained from other necklaces under rotation are considered the same)

We could stay that there are $n$ different ways to color the first bead, $n$ different ways to color the second, etcetera. But there are $5$ different rotations for each collar. Therefore the answer must be $\frac{n^5}{5}$, right? No, but close enough.

For instance, if all of the beads have only one color, then this necklace won't generate any other necklace under rotation. On the other hand, if a necklace has more than one color, then we affirm that it generates $5$ different necklaces (including itself). This is not an immediately obvious result so we shall prove it formally.

Indeed, let $(a_1,a_2,a_3,a_4,a_5)$ be a necklace, where $a_k$ is the color of the bead $k$. By assumption, there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$ (consider the indices modulo $5$). Suppose we rotate the necklace by $r$ places. We affirm that $(a_1,a_2,...,a_5)=(a_{1+r},a_{2+r},...,a_{5+r})$ if and only if $r$ is a multiple of $5$. Suppose it's not, and $a_{i}=a_{i+r}$ for all $i$. Inductively, $a_i=a_{i+mr}$, for $m\in\Bbb{Z}$. But since we're looking at the indices modulo $5$, let $m=(j-i)\cdot r^{-1}$ where $r^{-1}$ is the inverse of $r$ modulo $5$ (it exists since $5$ is prime and $5\nmid r$). Then $a_i=a_{i+mr}=a_{i+(j-i)r^{-1}r}=a_j$ for all $i,j$. But we're assuming there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$. Absurd.

Therefore the number of ways is actually $\frac{n^5-n}{5}+n$. The nice fact is that this number must be an integer (quite obviously), so in particular $\frac{n^5-n}{5}=\frac{n(n^4-1)}{5}$ is an integer. If $5\nmid n$, then $\gcd(5,n)=1$ since $5$ is prime, therefore $5\mid n^4-1\Rightarrow n^4\equiv 1 \pmod 5$. $\,\,\blacksquare$

Answer 9

Also, you can prove that $5|n^5-n$ by induction on $n$ and from this conclude the claim.

Answer 10

Fermat's little theorem provides an easy way.

But if you're not completely convinced, consider the squares modulo 5: $0, 1, 4, 4, 1$. Then the cubes: $0, 1, 3, 2, 4$. Then the fourth powers: $0, 1, 1, 1, 1$. Voila.

Answer 11

When $n\equiv 1\pmod{5}$, $n^4-1\equiv 1^4-1\equiv 0$.

When $n\equiv 2\pmod{5}$, $n^4-1\equiv 2^4-1=16-1=15\equiv 0$.

When $n\equiv 3\pmod{5}$, $n^4-1\equiv 3^4-1=81-1=80\equiv 0$.

When $n\equiv 4\pmod{5}$, $n^4-1\equiv 4^4-1=256-1=255\equiv 0$.

Answer 12

$$n^4-1 = (n-1)(n+1)(n^2+1)$$ The factors $n-1$ and $n+1$ take care of $n \equiv \pm 1 \mod 5$, while if $n \equiv \pm 2 \mod 5$, $n^2 + 1 \equiv 2^2 + 1 \equiv 0 \mod 5$.

Answer 13

One way to get the result is to apply the Euler's theorem: $\varphi(5)=4$ and $\gcd(n,5)=1$ so $n^{4}\equiv 1\pmod 5$.

Answer 14

Product of 5 consecutive integer numbers is of course divisible by $5$, so $5|(n-2)(n-1)n(n+1)(n+2)$. If $5\not|n,$ then from primality of $5$ we have \begin{align*} 5|(n-2)(n-1)(n+1)(n+2) & = (n^2-1)(n^2-4)\\ & = n^4 - 5n^2 +4\\ & = n^4 - 1 - 5(n^2 - 1) \end{align*} so $5|n^4-1$.

Answer 15

$n^{4}-1 = (n^{2}-1)(n^{2}+1)$ and (mod $5$) we have $n^{2}+1 \equiv n^{2} - 5n +6 = (n-2)(n-3),$ while also (mod $5$), we have $n^{2}-1 = (n-1)(n+1) \equiv (n-1)(n-4).$ Hence we have $(n^{4}-1) \equiv (n-1)(n-2)(n-3)(n-4) $ (mod $5$), and as long as $n$ is not divisible by $5$, one of $n-1,n-2,n-3,n-4$ is divisible by $5$.