Prove $(n^5-n)$ is divisible by 5 by induction.

Question

So I started with a base case $n = 1$. This yields $5|0$, which is true since zero is divisible by any non zero number. I let $n = k >= 1$ and let $5|A = (k^5-k)$. Now I want to show $5|B = [(k+1)^5-(k+1)]$ is true....

After that I get lost.

I was given a supplement that provides a similar example, but that confuses me as well.

Here it is if anyone wants to take a look at it:

Prove that for all n elements of N, $27|(10n + 18n - 1)$.

Proof: We use the method of mathematical induction. For $n = 1$, $10^1+18*1-1 = 27$. Since $27|27$, the statement is correct in this case.

Let $n = k = 1$ and let $27|A = 10k + 18k - 1$.

We wish to show that $27|B = 10k+1 + 18(k + 1) - 1 = 10k+1 + 18k + 17$.

Consider $C = B - 10A$ ***I don't understand why A is multiplied by 10. $= (10k+1 + 18k + 17) - (10k+1 + 180k - 10)$

$= -162k + 27 = 27(-6k + 1)$.

Then $27|C$, and $B = 10A+C$. Since $27|A$ (inductive hypothesis) and $27|C$, then $B$ is the sum of two addends each divisible by $27$. By Theorem 1 (iii), $27|B$, and the proof is complete.

Answer 1

Your induction hypothesis is that $5\mid k^5-k$, which means that $k^5-k=5n$ for some integer $n$. Now

$$\begin{align*} (k+1)^5-(k+1)&=\left(k^5+5k^4+10k^3+10k^2+5k+1\right)-(k+1)\\ &=k^5+5k^4+10k^3+10k^2+5k-k\\ &=(k^5-k)+5k^4+10k^3+10k^2+5k \end{align*}$$

can you see why that must be a multiple of $5$?

Answer 2

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

Thus it suffices to show that $\,5\,$ divides a product of $\,5\,$ consecutive integers. In fact, any sequence of $\,n\,$ consecutive naturals has an element divisible by $\,n\,$. This has a simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\,n,\,$ since it effectively replaces the old least element $\:\color{#C00}a\:$ by the new greatest element $\:\color{#C00}{a+n}$

$$\begin{array}{}& \color{#C00}a, &\!\!\!\! a+1, &\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1 & \\ \to & &\!\!\!\! a+1,&\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1, &\!\!\!\! \color{#C00}{a+n} \end{array}\qquad$$

Since $\: \color{#C00}{a\,\equiv\, a\!+\!n}\pmod n,\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\ 0,1,2,\ldots,n-1\ =\: $ all $ $ possible remainders mod $\,n.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\,n.$

Answer 3

You can also use decomposition over polynomials :$\quad\displaystyle \Pi_k(n)=k!\binom nk=\prod\limits_{i=0}^{k-1} (n-i)$ $\displaystyle n^5-n = 30\times\left[ 4\binom n5+8\binom n4+5\binom n3+\binom n2\right]$

Since the binomial coefficients are integers, you get the divisibility by $30$ as a result.

This method can be generalised to other problems of the style, "show that $m$ divides $P(n)$".

Answer 4

Since $$(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k - k$$, then $k^5 - k$ must be a multiple of $5$. And yes, it is by the fact that the unit digit of $k^5 - k$ will always be $0$. So the whole expression is really a multiple of $5$.

Answer 5

Using modular arithmetic: $$\begin{align}n&\equiv 0,1,2,3,4 \pmod 5 \\ n^5&\equiv 0,1,2,3,4 \pmod 5 \\ n^5-n &\equiv 0 \pmod 5.\end{align}$$

Answer 6

Also note that all the binomial coefficients $C(p, k) = \frac{p!}{k!(p-k)!}$ when $p$ is a prime are divisible by $p$ except for $k=0$ and $k=p$ (since $p$ is in the numerator but not in the denominator).

This is enough to show by induction that $n^p-n$ is divisible by $p$ for all positive integers $n$: True for $n=1$ and, for the induction,

$\begin{align} ((n+1)^p-(n+1))-(n^p-n) &=(n+1)^p-n^p-1\\ &= (n^p+(\text{terms divisible by p})+1)-n^p-1\\ &= (\text{terms divisible by p}) \end{align}$ .