Ok, take the quadratic polynomial

f(n) = n^2-n+41,

and notice that f(1)=41, f(2)=43, ... f(40)=1601 are *all* prime. There's a rather deep reason for this. (Thread)

The discriminant of the quadratic is

D = B^2-4AC = (-1)^2 - 4*41 = -163.

Here's a crazy fact about this number https://twitter.com/3blue1brown/status/1285019020379320320

2/ Let chi(n) = 0 if n divides D, 1 if n is a square mod D, and -1 otherwise (this is the "quadratic Dirichlet character" mod D). The first many primes are all non-squares mod D! Just like Riemann zeta, the L-function of chi

L(chi,s) = sum chi(n)/n^s

has an Euler product:

3/

L(chi,s) = prod (1-chi(p)/p^s)^-1

but the first many p's have chi(p)=-1. So this L-function is "trying very hard" (analytically) to pretend to have Euler product

prod (1+1/p^s)^-1,

which is the Euler product of zeta(2s)/zeta(s). (Why?...)

But look at what's happening:

4/ The function zeta(2s)/zeta(s) has a *zero* at s=1!!! Of course if L(chi,s) also had a zero at (or near) s=1, that would be a massive violation of GRH (this is the notorious Siegel zero issue...), and this *almost* happens here.

Digging deeper, the reason is that -163 is...

5/ the largest negative fundamental discriminant with *class number one*. By this I mean the following.

Let g be the binary quadratic form, say:

g(x,y) = 167x^2 - 169xy + 43y^2.

Check that it has the same discriminant,

169^2-4*167*43=-163.

It turns out that...

6/ an invertible integer linear change of variables turns g into

f(x,y) = x^2-xy+41y^2

(the homogenized version of f(x) above); indeed,

g(-x+2y, -2x+3y) = f(x,y),

and this transformation is invertible over the integers,

f(3x-2y,2x-y) = g(x,y).

Theorem: the same is true ...

7/ for *any* integer binary quadratic with discriminant -163 - they're *all* equivalent to f. Again, this is the last time this happens (it's not true for any more negative discriminants... not to be confused with positive discr, which have a completely different story...)

Let's

8/ look at what this means from the Class Number Formula. At s=1, we have:

L(chi,1) = pi h(D)/sqrt{|D|},

here h(D)=1 is the class number. If h(D) could be 1 infinitely often, 1/sqrt{|D|} -> 0, and so the L-function would be small and GRH would fail! (see: Deuring-Heilbronn)

9/ What does this have to do with n^2-n+41 being all primes early on?

Let w=(1+sqrt{-163})/2 so that the ring of integers of Q(sqrt{-163}) is Z[w]. The "norm form" of Z[w] is the binary quad f(x,y)=x^2-xy+41y^2; that is, if

z=x+y w \\in Z[w],

then (exercise)

|z|^2=f(x,y).

10/ So what happens if a (rational, that is, usual) prime p splits in Z[w] (and hence has Jacobi symbol +1... Why?)? That is, p is no longer prime because p=(x+yw)(x+y wbar). But that's the norm form, and p can't be expressed in any other way (class number one), so elementary...

11/ considerations force p>(|D|-1)/4=41. So all small primes are inert (as we saw before, the Jacobi symbol is all -1's early on). If some small value of |n+w|^2 = n^2-n+41 was *not* prime, it would give a non-trivial factorization in Z[w] involving small split primes, which

12/ don't exist!

Ok, this thread is already too long, so for the connection to e^{pi sqrt{163}}, let me just mumble something about the j-invariant function taking integer values at CM points of class number one imaginary quadratic fields, and having q-expansion dominated by

13/ the first term... See some more here:

https://en.wikipedia.org/wiki/Complex_multiplication#Singular_moduli

I'm afraid I'm not aware of a good reference (aimed at a lay audience) where these beautiful facts are laid bare!...