Youtube hearted comments of Thomas Morgan (@mrtthepianoman).

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
Very cool! When you gave your formula for generating each triple from the previous triple, that technique has a clear higher-dimensional analogue. If you generate n-tuples using this rule, I would imagine that would correspond to a golden n-dimensional hypercube, which would have a corresponding golden spiral. I’d be curious to know if these higher dimensional analogies had any interesting properties, applications, or divergences from their lower dimensional counterparts.
2
I like the coloring proof better
1
Is the 1/7 fraction in the Bizets formula a coincidence, or is that somehow related to the fact that these numbers are generated by the 7-gon? I would argue that the sqrt 5 that appears all over in Bizets formula for the Fibonacci sequence comes from the relation to the regular pentagon, so it seems like this would be the case here as well.
1