A long time ago (meaning 9 months), I discussed with @CasTur how to display visually a simple proof of Euler's Theorem, one of the most confusing results in elementary number theory. The statement of it is: If a and n are coprime, then a^phi(n) is congruent to 1 modulo n, where phi(n) is Euler's totient function, the number of integers below n coprime to it (the ones that don't share any factors with n). It's not too hard to prove, but to /intuitively/ prove it is another thing. Here, the n blue points distributed around the circle represent addition modulo n, where 0 is the top point. The points circled in red are invertible, meaning that they can be multiplied by another number modulo n to result in 1. These numbers also happen to be the only ones coprime to n (the proof is rather difficult)— and they form a group with the operation of multiplication. (A group is a set with an operation over it that is associative, with an identity element like 0 is for addition, and every element having an inverse.) This is a group because the product of two invertible elements is invertible, and the inverse of an invertible element is invertible as well! (A short proof of the previous sentence follows: if a is invertible, and b is invertible, then ac=1 and bd=1 for some c, d; ab is invertible because (ab)(cd)=1. If a is invertible, then ac=1 for some c; and c is the inverse of a, by definition, so c is invertible, because ca=1.) The group has order phi(n), obviously — and it is a cyclic group generated by n-1 (meaning that all of its elements are powers of n-1). It is a well-known fact from group theory that if k is the order of a cyclic group, the kth power of any element of that group is the identity element, which here is 1. Putting the pieces together, the phi(n)-th power of any number coprime to n is congruent to 1 modulo n, proving the theorem.
Delay is the time, in seconds, between changes in n. Thanks to George Andrews, @CasTur, and Charles Pinter.
