Greatest common divisor of two consequetive triangle numbers

Currently thinking of simple problems which showcase how the Euclidean algorithm can be applied to deducing more theoretical results. It’s a largely abandoned algorithm in the Swedish primary and secondary educational systems and I suspect it’s mostly because it is a)hard to teach other than out of habit and b) the computational skill of students is too unreliable for them to work enough problems in a short enough time for it to be taugh within a few classes. (and c) it’s not as useful as other things like elementary geometry)

The problem which I’m going to review here was just inspired by the pretty well known result that two consecutive Fibonacci numbers are relatively prime where a proof by the Euclidean algorithm is delightfully elegant but once you’ve shown that one in an educational setting I only think it fair to have available some problems showcasing the same method and which are of equivalent difficulty.

Finding the greatest common divisor of two triangle numbers only superficially applies the Euclidean algorithm and its division into two cases makes it necessary to identify the two cases first.

The two distinctive cases are when the index of the largest of two consecutive triangle numbers is even and the other when it is odd. A triangle number being alternately defined by

T_{n} = 1 + 2 + 3 + ... + n = \cfrac{n(n + 1)}{2}\qquad \text{or} \qquad T_{n} = T_{n-1} + n.

Case 1. The larger one has an odd index. T_{2n}, T_{2n +1}. Applying the Euclidean algorithm we have

T_{2n + 1} = T_{2n} + 2n + 1

T_{2n} = 2n(2n + 1)/2 = n(2n + 1) + 0

After two steps we get that the greatest common divisor was 2n + 1 and we summarize

\boxed{\gcd(T_{2n}, T_{2n + 1}) = 2n + 1}

Case 2: The larger one has an even index. T_{2n-1}, T_{2n} Applying the Euclidean algorithm again we have

T_{2n} = T_{2n-1} + 2n

T_{2n-1} = (2n-1)2n/2 = (2n-1)n = (n - 1)2n + n

(n -1)2n = 2n + 0

and we have thus found that if the larger triangle number has an even index then the greatest common divisor is n.

\boxed{\gcd(T_{2n-1}, T_{2n}) = n}

The connection between diameter and pi

So I was having the worst headache and lying in bed thinking about simple geometry problems waiting for the pain killer to kick in. One of the things I did was compute the perimeter/circumference of the Reuleaux triangle. Okay so that one’s parameter is composed of three sixths of a circle with radius twice the widthof the triangle so if the width of the triangle is D that’s O = 3 \cdot 2\pi D / 6 = \pi D. Similarly for all the other Reuleaux polygons the (n = 2k+1)-sided reuleaux polygon O = n \cdot 2\pi/(2n) D = \pi D. Huh… I thought. That’s kind of odd. All of these figures with constant widths have the same circumference formulas as the circle which of course has constant width called the diameter.

And looking up the wiki-article on shapes of constant widths and this is apparently a thing proven in 1860. All curves with constant width have a perimeters which is \pi times it’s width. and it’s called Barbier’s theorem.

I also get the feeling that this should be possible to be reworked into a pro-pi argument in the tau-vs-pi debate. Of course by the law of the equivalence between pi and tau implies it could be made into a pro-tau-argument as well but since I’ve never seen it come up in one of those debates I would like to see what happens when a tauist brings up that defining the circle constant via the diameter of a circle was a mistake since the radius is a more fundamental property and then have the piist counter that at least there is a whole class of objects with diameters and they all have perimeters which are \pi times the diameter : P