Calculus is the mathematics of approximations of continuous quantities; number theory asks exact problems about discrete quantities. These seem, therefore, to be unrelated studies.
Yet Kurt Hensel made a remarkable discovery: using ideas from calculus, one can better understand number theory problems. We're going to explain Hensel's wonderful discovery, by trying to solve polynomial equations in modular arithmetic. More particularly, we are going to try and find all integers \(x\) such that \[x^3 - 17x^2 + 12x + 16 \equiv 0 \pmod{3000}.\]
While this sort of problem -- solving a polynomial equation in modular arithmetic -- might seem a bit strange, it actually leads directly to the \emph{Langlands program} in modern number theory, as we'll comment a little about at the end.
The prime factorization of 3000 is \[3000 = 2^3 \times 3 \times 5^3.\]
The Chinese remainder theorem tells us that solving \[x^3 -17x^2 + 12x + 16 \equiv 0 \pmod{3000}\] is, therefore, equivalent to solving the \emph{three} different congruences \[x^3 - 17x^2 + 12x + 16 \equiv 0\pmod{2^3},\] \[x^3 - 17x^2 + 12x + 16 \equiv 0\pmod{3},\] \[x^3 - 17x^2 + 12x + 16 \equiv 0\pmod{125}.\]
It might seem like we've made our problem \emph{harder}, not easier: we used to have one equation, but now we have three! However, there is a sense in which our life has been made simpler: to solve \[x^3 - 17x^2 + 12x + 16 \equiv 0 \pmod{8},\] for example, we only need to plug in \(x = 0, 1, 2, 3, 4, 5, 6,\) and \(7.\) This is only eight values to try. If we simplify our polynomial to \[x^3 -17x^2 + 12x + 16 \equiv x^3 - x^2 + 4x \pmod{8},\] and plug in each of the possible values, we quickly find that the only solutions to our equation modulo 8 are \[x \equiv 0, 4 \pmod{8}.\]
Similarly, we can plug in \(x= 0, 1, 2\) to find that the only solution to \[x^3 - 17x^2 + 12x + 16 \equiv 0\pmod{3}\] is \(x \equiv 1 \pmod{3}.\)
Just by plugging in a few values, we managed to solve 2 out of 3 of our equations. Unfortunately, plugging in every value from \(x = 0\) to \(x = 124\) to solve \[x^3 - 17x^2 + 12x + 16 \equiv 0\pmod{125},\] seems like it would take much too long. Are there any quicker ways to proceed?
Recall \(125 = 5^3.\) So, if \(x\) is an integer making \[x^3 - 17x^2 + 12x + 16 \equiv 0 \pmod{125},\] then we also have \begin{equation} \label{eq52} x^3 - 17x^2 + 12x + 16 \equiv 0 \pmod{5}.\end{equation}
This equation is a little easier to solve: we can plug in \(x = 0, 1, 2, 3,\) and 4 to find that the only solutions to \autoref{eq52} is \[x \equiv 2 \pmod{5}.\]
When we plug in \(x = 2\) to our polynomial, we get \[2^3 - 17 \cdot 2^2 + 12 \cdot 2 + 16 = -20.\]
This number is indeed divisible by 5... but it is not a multiple of \(125.\) However, Hensel thought that \(-20\) is \emph{close} to being zero modulo 125, because \(125 = 5^3,\) and \(-20\) is at least divisible by 5 once, even if it isn't divisible by 5 three times.
On its own, this thought would probably not be so helpful. What made this idea so remarkable was what Hensel did \emph{next}. Before we can explain that, though, we need to take a brief digression to recall an insight from calculus.