[krackers]

>for any y >= n with f(y) = 0 (mod n), there's some x between 0 and n-1

There's a simpler way to see this, any such y can be represented as y = nk + x where i,j are divisor & remainder. Then f(y) = f(nk + x) = f(x) modulo n since by binomial theorem all other terms other than those with just x will be divisible by n.

[articulatepang]

Thank you and yes, I agree! It's neat to use the binomial theorem to see this, because that's the tool the article uses for the main trick/insight it's explaining.

[krackers]

>neat to use the binomial theorem to see this,

Yeah as far as I understand this basically _is_ where the connection to calculus comes in. As the article points out, it's very similar to computing the derivative via infinitessimals, e.g. (x+dx)^3 = x^3 + 3x^2 dx + O(dx^2), and from that you recover the term linear in dx as the derivative, 3x^2 . For the derivative you consider epsilon as a nilpotent where dx^2 = 0, and for hensel lifting lemma any term with a factor of p^2 gets zeroed (modulo p^2).

I'm just a layperson but I'm not really convinced that this is differential calculus in a meaningful sense. The other commenter mentioned "formal derivative" which seems fitting. There might be a connection to "umbral calculus" (something i know nothing about fwiw) though in the way you use the formal computations of derivatives for identities on polynomial equations, even without there being any differentials actually involved.