How to prove that $n^5 - n$ is a multiple of $5$?

Question

Hello I'm new to induction so please bare with me. For this problem I have to use induction to prove: For every integer $n\geq 1$, the number $n^5 − n$ is a multiple of $5$. Can someone please help me figure this out. I've tried to follow the examples in the book, but they were very confusing.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\Large\tt A "Non-Induction" Proof$:

Let $n = p + \delta$ where $p\in{\mathbb N}$. $10|p$. $0 \leq \delta \leq 9$. Then, $$ n^{5} = \pars{p + \delta}^{5} = \overbrace{% p^{5} + 5p^{4}\delta + 10p^{3}\delta^{2} + 10p^{2}\delta^{3} + 5p\delta^{4}} ^{\mbox{Obviosusly, it's a multiple of}\ 10}\ +\ \delta^{5} $$ $\delta^{5} \in \braces{\color{#00f}{\large 0},\color{#00f}{\large 1},3\color{#00f}{\large 2},24\color{#00f}{\large 3},102\color{#00f}{\large 4},312\color{#00f}{\large 5},777\color{#00f}{\large 6},1680\color{#00f}{\large 7},3276\color{#00f}{\large 8},5904\color{#00f}{\large 9}}$ such that the last digit of $n^{5}$ is equal to $\delta$. Then, the last digits of $n$ and $n^{5}$ are equal which means that the last digit of $n^{5} - n$ is zero and $\color{#00f}{\large 5 | \pars{n^{5} - n}}$.

Answer 2

Clearly for $n = 1$: $1^5 - 1 = $ which is a multiple of $5$

Induction Hypothesis $5$ divides $n^5 - n$. We need to show: $5$ divides $(n+1)^5 - (n+1)$.

Now, $(n+1)^5 - (n+1)$ = $n^5 + 5n^4 + 10n^3 + 10n^2 + 5n +1-n -1$. So $5$ divides $(n+1)^5 - (n+1)$.

So by Principle of Mathematical Induction $5$ divides $n^5 - n$ for all $n\in \mathbb{N}$.

Answer 3

Base case, $n=1$: $1^5-1=0$ is divisible by 5.

Hypothesis: $\exists n \geq 1: 5|(n^5-n)$

Step: $$(n+1)^5-(n+1)= (\sum_{k=0}^5\binom{5}{k}n^k) - (n+1) \\ = 1+ 5n+ 10n^2 + 10n^3+5n^4+n^5-n-1=5(n+2n+2n^2+n^4) + n^5-n$$ Now by the hypothesis $5|(n^5-n)$, and clearly $5|[5(n+2n+2n^2+n^4)]$, and therefore: $$ 5|[(n+1)^5-(n+1)]$$

This completes the proof by induction and it follows that: $$\forall n \geq 1: 5|(n^5-n)$$

If you don't believe in induction (I don't believe in induction, because it has nothing to do with belief ;-): $$n^5-n = n(n^4-1) = n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1)$$ Now, if: $$(n-1 \equiv 0 \bmod 5) \oplus (n \equiv 0 \bmod 5) \oplus (n+1 \equiv 0 \bmod 5)$$ We are done. If not, then: $$(n-2 \equiv 0 \bmod 5) \oplus (n+2 \equiv 0 \bmod 5)$$ Must hold. Now notice that $(n-2)(n+2)=n^2-4 \equiv n^2+1\bmod 5$.

So in this case, using modular multiplication rules, $n^2+1 \equiv 0 \bmod 5$ must hold.

So there's no escape! $5|(n^5-n)$.

Answer 4

I would like to show another approach without induction method

We know that

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

Now Let's focus on another sequence

$ n>2$ and $k>0$

$$(n-2)(n-1)n(n+1)(n+2)= 5.k$$

Because if we have 5 numbers sequence and always we will have a number in the sequence such form of $5.j$.

Examples:

$3.4.5.6.7$

or $4.5.6.7.8$

or
$5.6.7.8.9$

or $6.7.8.9.10$

another sequence also must be such form of

$$(n-1)n(n+1).5= 5.m$$

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

To sum up both terms , you can get

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

$$(n-1)n(n+1)(n^2+1)=n^5 - n= 5.(k+m)$$

Answer 5

Proof:

Base case $n = 1: 1^5 - 1 = 0 = 0*5$ which is a multiple of $5 ( k = 0 )$.

Assume $5$ divides $n^5 - n$. We need to show: $5$ divides $(n+1)^5 - (n+1)$.

$(n+1)^5 - (n+1) = n^5 + 5k + 1 - (n+1) = 5k + n^5 - n$. We have: $5$ divides $5k$, and from induction step: $5$ divides $n^5 - n$. So $5$ divides $(n+1)^5 - (n+1)$.

So: $5$ divides $n^5 - n$ for all $n$.