This is a nice post; thanks for writing it! (A minor thing: the margin notes completely disappear on mobile, i.e. at width of 760px or less.)
You probably know this already, but I'd think of this the following way. There are two main mathematical ideas involved here:
• The first, easy to underrate because it can seem "obvious", is the Chinese remainder theorem. This, for instance, here implies that any function of the quantities (x mod 3) and (x mod 5) can be rewritten as a function of just (x mod 15). So if you used just this one idea and not the next one, you could implement FizzBuzz as
lambda n: ['FizzBuzz', n, n, 'Fizz', n, 'Buzz', 'Fizz', n, n, 'Fizz', 'Buzz', n, 'Fizz', n, n][n % 15]
• The second is Fermat's little theorem. It says (x^(p-1) mod p) = [x is not a multiple of p], where the notation […] is Iverson bracket, i.e. 1 or 0 depending on whether the condition is true or not. So the question of whether x is a multiple of 5 or not is equivalent to whether x^4 mod 5 is 0 or 1. This just gives us a convenient way of restating the divisibility condition.
To get from Fermat's little theorem to Euler's theorem (or to be pedantic, Carmichael's theorem, as you're not using φ(15) which is technically 2*4 = 8, but rather using lcm(2, 4)=4: https://en.wikipedia.org/w/index.php?title=Carmichael_functi... ) is itself an application of the Chinese remainder theorem, which is why I think it's important and mentioned it first.
And putting these two ideas together gives the function in your post. Namely: the FizzBuzz you want is a function of [x is a multiple of 3] and [x is a multiple of 5], so you can (using the second idea) rewrite it as a function of (x^2 mod 3) and (x^4 mod 5), and put them together (using the CRT) as a function of (x^4 mod 15).
Coincidentally, both the ideas here are connected to Lagrange: the Chinese remainder theorem is the same kind of thing as the Lagrange interpolation formula (https://artofproblemsolving.com/community/c1157h990758_the_c...), and Fermat's/Euler's/Carmichael's theorem is the same kind of thing as Lagrange's theorem in group theory.
Re the margin notes, the footnote numbers are clickable to toggle them inline when the width is too small... I should add a bit of color or underline to them in the css so that this is easier to intuit. :/
Ah, clicking the footnote numbers worked for footnotes 1 to 4, but not for 5 and 6.
BTW, for the question “Where do the constant values 0, 6, 10, and 1 come from?”, though it's implicit in the post, it is useful to note explicitly that as x^4 takes only the values 0 or 1 either mod 3 or mod 5, these four possibilities—namely (0,0), (0,1), (1,0), (1,1)—precisely account for those values:
(0 mod 3) and (0 mod 5) ⇔ (0 mod 15)
(0 mod 3) and (1 mod 5) ⇔ (6 mod 15)
(1 mod 3) and (0 mod 5) ⇔ (10 mod 15)
(1 mod 3) and (1 mod 5) ⇔ (1 mod 15)
This would also simplify the post considerably maybe, as much of the algebra wouldn't be needed.
You probably know this already, but I'd think of this the following way. There are two main mathematical ideas involved here:
• The first, easy to underrate because it can seem "obvious", is the Chinese remainder theorem. This, for instance, here implies that any function of the quantities (x mod 3) and (x mod 5) can be rewritten as a function of just (x mod 15). So if you used just this one idea and not the next one, you could implement FizzBuzz as
• The second is Fermat's little theorem. It says (x^(p-1) mod p) = [x is not a multiple of p], where the notation […] is Iverson bracket, i.e. 1 or 0 depending on whether the condition is true or not. So the question of whether x is a multiple of 5 or not is equivalent to whether x^4 mod 5 is 0 or 1. This just gives us a convenient way of restating the divisibility condition.To get from Fermat's little theorem to Euler's theorem (or to be pedantic, Carmichael's theorem, as you're not using φ(15) which is technically 2*4 = 8, but rather using lcm(2, 4)=4: https://en.wikipedia.org/w/index.php?title=Carmichael_functi... ) is itself an application of the Chinese remainder theorem, which is why I think it's important and mentioned it first.
And putting these two ideas together gives the function in your post. Namely: the FizzBuzz you want is a function of [x is a multiple of 3] and [x is a multiple of 5], so you can (using the second idea) rewrite it as a function of (x^2 mod 3) and (x^4 mod 5), and put them together (using the CRT) as a function of (x^4 mod 15).
Coincidentally, both the ideas here are connected to Lagrange: the Chinese remainder theorem is the same kind of thing as the Lagrange interpolation formula (https://artofproblemsolving.com/community/c1157h990758_the_c...), and Fermat's/Euler's/Carmichael's theorem is the same kind of thing as Lagrange's theorem in group theory.