Comments by "Thomas Morgan" (@mrtthepianoman) on "The Film with the Most Maths" video.

  1. Here's my (arguably) simpler proof of the first problem. It's a proof by induction. Base case: the pair 1, 2 works (easy to find). Induction step: assume we have found a pair a, b such that a < b and a | b^2 +1 and b | a^2 +1. Then we can write b^2 + 1 = a * c for some positive integer c. Then it is clear that c | b^2 + 1. Thus it suffices to show that b | c^2 + 1. But c = (b^2 + 1) / a, so c^2 +1 = (b^4 + 2b^2 + 1) / a^2 + 1 = (b^4 + 2b^2 + a^2 + 1) / a^2. We know b | a^2 + 1, so b divides the numerator. But if b | a^2 + 1, then b is relatively prime to a^2 (the denominator). It follows that b divides the quotient, and we are done. Thus, starting from any pair (a, b) we can find, we can generate an infinite sequence as follows. Start with (a, b). b^2 + 1 = a * c gives us the pair (b, c). c^2 + 1 = b * d gives us the pair (c, d). And so on. This proof is nice because it does not require any knowledge about properties of Fibonacci numbers. And it follows easily from the pattern of the first few examples that you found (and they were the first few examples I found as well).
    29
  2. I should also add that c > b, otherwise there is no guarantee that my construction would go in a circle. That follows from the fact that b > a, which implies c = (b^2 + 1)/a > (a^2 + 1)/a > a^2/a = a
    5