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I will share a little proof I found that the number of loops in a Celtic knot of dimension a x b is the greatest common divisor of a and b. The drawing below by @anniek_p of a 7 x 14 Celtic knot illustrates this. It has 7 loops, each in a different color: #mathartchallenge
Lemma 1: An n x n Celtic knot has exactly n loops.
Proof: Just look at it! The loops organize as tilted rectangles. Notice also that each loop goes around exactly once - this will be useful in a minute (elegant drawing by @teafairy79 for n = 4).
Proof: Just look at it! The loops organize as tilted rectangles. Notice also that each loop goes around exactly once - this will be useful in a minute (elegant drawing by @teafairy79 for n = 4).
Lemma 2: If a>b, an a x b knot has as many loops as an (a-b) x b one.
Proof: Draw out a b x b square. Each loop goes around just once there, so we can “shrink it” (as pictured). This changes the knot structure but not the number of loops (original drawing by @mtrushkowsky).
Proof: Draw out a b x b square. Each loop goes around just once there, so we can “shrink it” (as pictured). This changes the knot structure but not the number of loops (original drawing by @mtrushkowsky).
If we do this for each loop in the b x b square what is left is an (a-b) x b Celtic knot, which we conclude has the same number of loops as the original one.
This suffices to finish the proof: start with an a x b knot and repeatedly subtract the small dimension from the large until we get equal numbers. By Lemmas 1 and 2, this is the original number of loops. But this is the Euclidean algorithm, so this number is the gcd of a and b!
Thanks again to @anniek_p for doing the wonderful #mathartchallenge and to @mtrushkowsky who wrote this post that got me thinking about this stuff https://twitter.com/mtrushkowsky/status/1247616970616221696.
