[MIT] Putnam Study Group

30 days Spent on Solving MIT problem

This is one of the hardest problems from MIT Putnam practice group. And It really is.

Here’s some opinion that i had after solving this problem

Would i be able to solve it in Putnam Exam??

Heck no, I spent 5 days solving this, with help of Internet. I think i wont be even be able to get into the exam room because i would end up just sleeping the whole day and rott in bed too lol.

Well with this idea in mind, lets try to get into this problem.

[5] Let $\left(\frac{n}{7}\right)$ denote the Legendre symbol. Specifically,

Show that

The main point of this exercise is to give an example of an explicit identity that can be computed to any degree of accuracy but which is only conjecturally true.

Problems are marked by the following difficulty ratings.

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[1] Easy. most students should be able to sovle it.
2

3
[2] Somewhat difficult or tricky. Many students should be able to sovle it.
4

5
[3] Difficult. only a few students should be able to solve it, but those who like a challenge
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might want to give it a try.
7

8
[4] Horrendously difficult. We don't really expect anoyone to solve it, but those who like a
9
challenge might want to give it a try.
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[5] Unsolved.

we can change the tan into sin yes yes yes and then we can do gaussian sum. Clausian derivative???? We can use Clausen-dilog identity

These were the initial thought, but let’s dive deeper!

Fourier Series LHS

We will takle the LHS first.

For LHS we can clean up the log using

and then we can expand with

integrate term-by-term over $[\pi / 3, \pi / 2]$ to get a series

Set $\beta = arctan \sqrt{7}$

Using

We can see that $\beta \in (\pi / 3, \pi / 2)$ so integrand has a logarithmic singularity at $t = \beta$ but this is integrable.

Each subinterval $[\pi/3, \beta - \varepsilon]$ and $[\beta + \varepsilon, \pi / 2]$ both $t + \beta$ and $\left| t - \beta \right|$ lies on a compact subset of $(0, \pi)$ which is the region where the Fourier series for $\log(2 \sin x)$ is convergent uniformly.

Fourier Expansion

like we mentioned above, subintervals lies on a compact subset which is also the region where Fourier series is uniformly convergent. So we can use $\log (2 \sin x)$ to integrate term by term.

For $0 \lt x \lt \pi$ we have this famous expansion

So,

the $\log 2$ cancels out and then we use this identity

To our case which A is $2n(t + \beta)$ and B is $2n(t - \beta)$ to get

And

![[Pasted image 20251223040308.png]]

We can and then integrate over $[ \pi / 3, \pi / 2]$ and term by term we can justify using the uniform convergence on the split intervals and letting $\varepsilon \rightarrow 0$, this dominated convergence works because the partial sums are bounded and also because original integrand was integrable.

Integrate!

So you get this series for the integral

This was where i got stuck… But this problem was in my dreams and took quite a while to get a hint from llm. But this was the place where we can change into Calusen perspective

Combination of Clausen functions, which can be defined by

We can use this product to sum formulas and package the integral we got above above.

Gauss Sums n Dilogarithms

So we will now tackle right hand side

This is a Dirichlet L value. To do this, we have to change this quite “bumpy” function $\mathcal{X}(n)$ to smooth exponentials, and we can do this using Gauss sum.

Let $\zeta = e^{2 \pi i / 7}$ We can define the Gauss sum

We need two specific facts from analytic number theory to finish off. so I will bring out the “Evaluation of the Quadratic Gauss Sum” by Murty & Pathak for these

Fact 1 Twist Identity

In Murty-Pathak, they define the quadratic Gauss sum $G(2) = \sum^{n-1}_{j=0} e^{2 \pi i j^2 /2}$. They show that for a prime $p$. this can be also rewritten as a sum involving Legendre symbols, which is equal to our $\tau(\mathcal{X})$ The Lemma 4.1 shows this

This makes us easier to allow to write $\mathcal{X}(n)$ as Fourier coefficient

Fact 2 Evaluation of the Sum

We need to get the value of $\tau(\mathcal{X})$. The Theorm 1.1 in Murty-Pathak states that for $n = 3 (\mod 4)$ the quadratic Gauss sum is purely imaginary with magnitude of $\sqrt{n}$ so for $n=7$

Since we know these facts, we can finally attempt to turn this series into dilogarithms

We can substitute $3$ into our sum $S$

$S = \sum_{n \ge 1} \frac{1}{n^2} \left[ \frac{1}{\tau\mathcal(X)} \sum^6_{a=1} \mathcal{X}(a)\zeta^{an} \right]$

and then we swap the summation order to find our dilogarithm we can use our handy Spence function $\operatorname{Li}_2(z) = \sum z^n/n^2$ $S = \frac{1}{\tau(\chi)} \sum_{a=1}^{6} \chi(a) \sum_{n\ge1} \frac{(\zeta^a)^n}{n^2} = \frac{1}{i\sqrt{7}} \sum_{a=1}^{6} \chi(a) \operatorname{Li}_2(\zeta^a).$

Dilog $\to$ Clausen

To get back to real numbers, pair the terms $a$ and $7-a$ Note that we have $\chi(7-a) = \chi(-a) = -\chi(a)$ and $\zeta^{7-a} = \bar{\zeta}^a$

Summing for $a=1, 2, 3$ and substituting into our expression for $S$

The $i$ and $1/i$ cancel perfectly so we get this hot answer

Calculation in Python

Too lazy to do calculation by hand

import math 
 
def legendreballs(n): 
	r = n % 7 
	if r == 0: return 0 
	return 1 if r in (1,2,4) else -1
	
def RHS(N=200000):
	return sum(legendreballs(n)/(n*n) for n in range (1, N+1))
	
 
# numerical quadrature 
def LHS(M=200000): 
	a, b = math.pi/3, math.pi/2 
	h = (b-a)/M 
	s = 0.0 
	
	for k in range(M+1): 
		t = a + k*h
		if k == 0 or k == M: w = 0.5 
		else: w = 1.0 
		val = abs((math.tan(t) + math.sqrt(7.0))/(math.tan(t)-math.sqrt(7.0)))
		s += w*math.log(val) 
	integral = h*s 
	return (24.0/(7.0*math.sqrt(7.0)))*integral
 
print(LHS(), RHS()) 

uhh this will gives us

1
1.1519281128932546 1.151925470544491

hm can’t we use mpmath and split at cusp?

import mpmath as mp
mp.mp.dps = 50 
 
chi = {1:1, 2:1, 3:-1, 4:1, 5:-1, 6:-1} 
 
# Left hand side where we inveral at singular point 
a, b = mp.pi/3, mp.pi/2
beta = mp.atan(mp.sqrt(7))
 
def integrand(t):
    return mp.log(abs((mp.tan(t) + mp.sqrt(7))/(mp.tan(t) - mp.sqrt(7))))
 
# Endpoint singularities are handled by mpmath 
# so uhhh we need to split 
I = mp.quad(integrand, [a, beta, b]) 
LHS = (24/(7*mp.sqrt(7))) * I 
 
# RHS using trigamma balls 
RHS = (1/49) * sum(chi[r] * mp.polygamma(1, mp.mpf(r)/7) for r in range(1,7))
 
print("LHS", LHS)
print("RHS", RHS)
print("abs diff", abs(LHS-RHS))

Much precise way on calculating

1
LHS 1.1519254705444910471016923973205499647978214046866
2
RHS 1.1519254705444909551811732061792532977684945907902
3
abs diff 9.19205191911412966670293268138964

Well, we can solve it numerical, but we are not able to use this in real exam environment which is pain in the ass, so if we want to solve this using proofs, we have to use arithmetic geometry. The reason why equality hold is that both sides of the equation (LHS and RHS) represents the invariance of the imaginary quadratic field $K = \mathbb{Q} (\sqrt{7})$

So we have to consider the Dedekind Zeta function where for the field $K = \mathbb{Q} (\sqrt{7})$ can be denoted as $\zeta_K(s)$. For the imaginary field, this function factors into a standard Riemann zeta function and the Dirichlet L function where it is associated with

and if we evaluate this with $s = 2$ we can get

So doing this our RHS series are just basically $\zeta_K(2)$ with a factor that can be up to $\pi^2 / 6$

To explain this better, we can actually use Borel’s Regulator theorm, which we will be using this book.

Borel’s Regulator Theorm, in algebraic number theory that relates the ranks of higher algebraic K-groups of the ring of integers of a number field to the orders of vanishing of the associated Dedekind zeta function at negative odd integers

Aaaaand especially for $K_3$, it states that for a number field $K$, the value $\zeta_K(2)$ is proportional to the Regulator $R_2(K)$ divided by the square root of the discriminant.

In the context of imaginary quadratic fields, the Regulator $R_2(K)$ is calculated by summing the Bloch-Wigner Dilogarithm evaluated at specific algebraic numbers within the field. The Bloch-Wigner function is essentially the “imaginary part” of the Dilogarithm which, as we saw at first part is exactly the Clausen function ($\mathrm{Cl}_2$)

1. RHS Calculates $\zeta_K(2)$ using the analytic definition
2. LHS Integral calculates $\zeta_K(2)$ using the geometric definition (the Regulator). The angle $\beta = \arctan(\sqrt{7})$ appears naturally because it relates to the argument of the fundamental algebraic integer in this field $$\alpha = \frac{1 + i\sqrt{7}}{2} \implies \tan(\arg(\alpha)) = \sqrt{7}$$

This explains why geometry matches the arithmetic, finishing our proof.

We basically verified steep theorm of algebraic K-theory! Not bad for a day’s work, it was worth it.