John Baez on the number 24
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when discussing why 24 is special, especially the fact that 1"² + ... + 24² is a perfect square, and as background for questions about sums of squares.
Recommended by 1 people across 1 different months, 2016 to 2018.
What it is
A one-hour public lecture from the 2008 Rankin Lectures series in which mathematical physicist John Baez explains why 12 and 24 recur throughout mathematics. It links Euler's 1+2+3+... = -1/12, the Riemann zeta function, bosonic strings in 26 dimensions, the Leech lattice, the Monster group and modular forms.
What people said
Recommends Stefan Gruenwald · 2016-08 · Mathematics Stack Exchange · on “Cool property of the number $24$”
24 is a very special integer number in many regards. John Baez has a nice pdf file about it: [ [LINK] ], and there is a very good Youtube video you can watch: [this video]
Cites Maik Pickl · 2016-10 · Mathematics Stack Exchange · on “When is the sum of the first $n$ squares a square?”
After watching John Baez talking about the number 24 [this video] I got interested in the question: For which $n \in \mathbb{N}$ is there a number $m \in \mathbb{N}$ s.t. $$ \sum_{k=1}^nk^2=m^2 $$ Or to put it in plain english, when is the sum of the first $n$ square numbers again a square. The surprising answer to this is that this is only the case for $n=0,1,24$ and you proof this (according to him) with the formula $$ \sum_{k=1}^nk^2=\dfrac{n(n+1)(2n+1)}{6} $$ and the machinery of elliptic curves. My guess is that you probably want to proof that the elliptic curve $E$ given by the above equation satisfies $E(\mathbb{Q})=\mathbb{Z}_3$ (not sure here, but something along these lines). My main qu
Cites Jordan Blevins · 2018-03 · Mathematics Stack Exchange · on “The Basel Problem, Cannonball Problem & 24-ness?”
Setting the finite sum of squares equal to a square number yields only one non-trivial solution, when $n= 24$; the sum becomes $4900$, which is $70^2$ $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ [this video] The infinite sum of reciprocal squares is equivalent to evaluating the Riemann Zeta Function at $s=2$. $Zeta(2) = \frac{\pi^2} 6$ [LINK] If you replace $\pi$ with $\tau (\tau = 2*\pi = 6.2931...)$ we can see that $Zeta(2)$ equals $\frac {(\tau^2)}{24}$ $24$ appears in both summations of squares. One way is a finite sum starting at the apex of a square pyramid toward an infinitely far away base, and the other problem involves starting at the base and going up towards an infinitely far away