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.
Best Answer
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 $5$?