[HN Gopher] Using calculus to do number theory
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Using calculus to do number theory
Author : cpp_frog
Score : 104 points
Date : 2026-03-16 14:12 UTC (2 days ago)
(HTM) web link (hidden-phenomena.com)
(TXT) w3m dump (hidden-phenomena.com)
| joshuaissac wrote:
| The mathematical field of tackling number theory problems in this
| way is called analytic number theory.
|
| https://en.wikipedia.org/wiki/Analytic_number_theory
|
| The prime number theorem, on how prime numbers are distributed
| amongst the integers, was first proved using analytic techniques.
| arch1t3cht wrote:
| Analytic number theory exists and involves calculus, but it's
| not what the linked post is about. The article talks about
| Hensel's lemma, which is a purely algebraic statement with a
| purely algebraic proof, which, however, is _inspired_ by
| techniques from calculus. This is typically still categorized
| as algebraic number theory.
| jjgreen wrote:
| Get a load of number theorists in a room and there will
| always be a fight between the analytic and algebraic.
| adampunk wrote:
| It's delightful (and unsurprising) that Newton's method shows up
| as the main bridge.
| pfortuny wrote:
| In some sense the title is a bit misleading (one is led to think
| of Analytic Number Theory). I'd rather use the title "Using
| differentials to...", which is more precise, as there is not
| exactly any "calculus" going on but there is indeed differential
| algebra and differential "number theory", so to speak.
|
| But great and elegant article. Thanks.
| measurablefunc wrote:
| The technical term is "formal derivative" b/c there are no
| limits involved, it's basically a rewrite rule for changing x^n
| to nx^(n-1).
| pfortuny wrote:
| I know, yes, but did not want to digress. Thanks though.
| Actually, it is the "extension" to K[e] with e^2=0, as
| "usual".
| measurablefunc wrote:
| The infinitesimally thickened point.
| ne38 wrote:
| Actually it is Calculus in p-adic numbers!
| pfortuny wrote:
| Mmmmhhhh, sure? Because p-adic numbers have characteristic 0,
| AFAIK.
| jjgreen wrote:
| If author is following this: minor typo in "we use the seem
| approximation trick again": seem -> same
| NooneAtAll3 wrote:
| author was so excited to point at calculus, that he forgot to
| derive actual mod3000 answer from chinese remainders...
|
| and it seems much more intuitive for me to see this technique as
| "find x mod p^n, then apply x->ans+p*x transformation and do
| everything at mod p^n+1" - and to derive that it results in
| derivatives from that
| articulatepang wrote:
| I love complex analysis, and that's the branch of calculus that
| is most associated with number theory. For example, it was
| critical in the original proof of the prime number theorem and
| Dirichlet's theorem on primes in arithmetic progressions. Today,
| all kinds of number theoretic functions are studied using complex
| analysis, like the famous Riemann zeta function, Dirichlet
| L-functions, theta functions, and so on.
|
| So that's what I expected this to be about. And it was super fun
| to see that it was actually not! Definitely learned something
| today.
|
| I had to think about the following sentence for a bit: "We know
| that any integer x obeying f(x)[?]0 (mod 125) must obey x[?]2(mod
| 5)."
|
| My explanation (it's basically all in the article, but here I
| spell it out): If f(x) = 0 (mod 125), then 125 divides f(x).
| Since 125 = 5^3, it must be that 5 also divides f(x). The only
| way for 5 to divide f(x) is for x = 2 (mod 5), by brute-forcing
| all solutions to f(x) = 0 (mod 5). Therefore for f(x) = 0 (mod
| 125), x = 2 (mod 5).
|
| It's also worth saying why we only need to check all integers
| between 0 and n-1 when solving an equation mod n:
|
| Suppose that for some integer y, f(y) = 0 (mod n) but y >= n or y
| < 0. Then for some x between 0 and n-1 (inclusive),
|
| x = y (mod n).
|
| Since the function f is a polynomial with integer coefficients,
| evaluating f on an integer involves only multiplication and
| addition by integers. Some crucial facts about congruences:
|
| If x = y (mod n) then a + x = a + y (mod n) for any integer a. If
| x = y (mod n) then ax = ay (mod n) for any integer a. If x = y
| (mod n) then x^2 = y^2 (mod n), and similarly other integer
| powers.
|
| From these we conclude that if x = y (mod n), then f(x) = f(y)
| (mod n) for any polynomial f.
|
| So, for any y >= n with f(y) = 0 (mod n), there's some x between
| 0 and n-1 (inclusive) that also satisfies f(x) = 0 (mod n); in
| fact it's whatever integer in that range that y is congruent to.
| So we need only check integers in that range to find all possible
| solutions to f(x) = 0 (mod n).
|
| Forgive me for the long explanation for what are of course
| elementary facts in number theory! I'm rusty on number theory so
| I had to explicitly work them out, so I figured maybe someone
| else might also benefit.
| krackers wrote:
| >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 = n _k + x where i,j are divisor & remainder.
| Then f(y) = f(n_k + x) = f(x) modulo n since by binomial
| theorem all other terms other than those with just x will be
| divisible by n.
| paulpauper wrote:
| "Appendix: The Langlands program"
|
| no offense but this doesn't even come close to describing it at
| all
| obastani wrote:
| This idea leads to the p-adic numbers:
|
| https://en.wikipedia.org/wiki/P-adic_number
| nomemory wrote:
| Blog looks promising. Is there a RSS feed associated with it?
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