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.