Matematiska gåtor sökes.

Solutions:
We First like to prove that if there are n blue-eyed people on the island, then they would all leave
the island on n-th day. If we can prove that, then surely we prove the original proposition.
We induct over n, the number of blue-eyed people.
Base case: n = 1. Well, the one lone blue-eyer would look at everyone else, realize that he/she is
the only person with blue-eye and hence leave on the First day.
Now assume that n blue-eyed people will all leave on the n-th day. Now suppose there are n + 1
blue-eyed people on the island. Let Min be a blue-eyed person. He would then look around the
island and see n blue-eyed people. If he himself has brown-eyed people, then all n people would leave
on the n-th day by inductive hypothesis. But this won't happened because every other blue-eyed
people would see n blue-eyes also. Thus, on the n+1 day, upon seeing the island still populated with
blue-eyers, Min would realize that he himself has blue-eye and thus leave the island on that day.
Since the situtation is symmetric, every other blue-eyed person would reach the same realization
and all leave the island on the n + 1 day.

a(n)=n^5-n
Induktion: a(1)=1-1=0, delbart med 5. Anta att a(n) är delbart med 5, och betrakta a(n+1):
a(n+1)=(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1=a(n)+5(n^4+2n^3+2n^2+n), vilket alltså är delbart med 5 och pga induktionsprincipen gäller det för alla n.

alternativ lösning:
om n=±1mod5, är n^5=±1mod5, är n=±2mod5, är n^5mod5=(±2)^5mod5=±32mod5=±2mod5, är n=3mod5, är n^5mod5=(3^5)mod5=243mod5=3mod5. alltså är nmod5=n^5mod5. betrakta nu a(n)mod5:
a(n) mod 5=(n^5-n)mod5=n^5mod5-nmod5=nmod5-nmod5=0 mod5
alltså 5|a(n).

n^5-n = n(n^4-1) = n(n^2+1)(n^2-1) = n(n^2+1)(n+1)(n-1) = (n^2+1)(n+1)(n-1)n
Subtrahera och addera med 5(n+1)(n-1)n:
(n^2+1)(n+1)(n-1)n - 5(n+1)(n-1)n + 5(n+1)(n-1)n
= (n^2-4)(n+1)(n-1)n + 5(n+1)(n-1)n =
(n+2)(n-2)(n+1)(n-1)n + 5(n+1)(n-1)n

Eftersom högra termen ovan har faktorn 5, och eftersom vänstra termen har faktorerna (n-2), (n-1), n, (n+1) och (n+5), alltså fem tal efter varandra så måste båda termerna vara delbara med 5.