Question

prove that n^5-n is divisble by 5.​

Answer 1

Answer:

$\huge\purple {\mathfrak{Bonjour Mate!}}$

So I started with a base case n=1n=1. This yields 5|05|0, which is true since zero is divisible by any non zero number. I let n=k>=1n=k>=1 and let 5|A=(k5−k)5|A=(k5−k). Now I want to show 5|B=[(k+1)5−(k+1)]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)27|(10n+18n−1).

Proof: We use the method of mathematical induction. For n=1n=1, 101+18∗1−1=27101+18∗1−1=27. Since 27|2727|27, the statement is correct in this case.

Let n=k=1n=k=1 and let 27|A=10k+18k−127|A=10k+18k−1.

We wish to show that 27|B=10k+1+18(k+1)−1=10k+1+18k+1727|B=10k+1+18(k+1)−1=10k+1+18k+17.

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

=−162k+27=27(−6k+1)=−162k+27=27(−6k+1).

Then 27|C27|C, and B=10A+CB=10A+C. Since 27|A27|A (inductive hypothesis) and 27|C27|C, then BB is the sum of two addends each divisible by 2727. By Theorem 1 (iii), 27|B27|B, and the proof is complete.