# Category Archives: Math.NT

## Talk: Finding Congruent Numbers, Arithmetic Progressions of Squares, and Triangles

Here are some notes for my talk Finding Congruent Numbers, Arithmetic Progressions of Squares, and Triangles (an invitation to analytic number theory), which I’m giving on Tuesday 26 February at Macalester College.

The slides for my talk are available here.

The overarching idea of the talk is to explore the deep relationship between

1. right triangles with rational side lengths and area $n$,
2. three-term arithmetic progressions of squares with common difference $n$, and
3. rational points on the elliptic curve $Y^2 = X^3 – n^2 X$.

If one of these exist, then all three exist, and in fact there are one-to-one correspondences between each of them. Such an $n$ is called a congruent number.

By understanding this relationship, we also describe the ideas and results in the paper A Shifted Sum for the Congruent Number Problem, which I wrote jointly with Tom Hulse, Chan Ieong Kuan, and Alex Walker.

Towards the end of the talk, I say that in practice, the best way to decide if a (reasonably sized) number is congruent is through elliptic curves. Given a computer, we can investigate whether the number $n$ is congruent through a computer algebra system like sage.1

For the rest of this note, I’ll describe how one can use sage to determine whether a number is congruent, and how to use sage to add points on elliptic curves to generate more triangles corresponding to a particular congruent number.

Firstly, one needs access to sage. It’s free to install, but it’s quite large. The easiest way to begin using sage immediately is to use cocalc.com,  a free interface to sage (and other tools) that was created by William Stein, who also created sage.

In a sage session, we can create an elliptic curve through


> E6 = EllipticCurve([-36, 0])
> E6
Elliptic Curve defined by y^2 = x^3 - 36*x over Rational Field


More generally, to create the curve corresponding to whether or not $n$ is congruent, you can use


> n = 6   # (or anything you want)
> E = EllipticCurve([-n**2, 0])


We can ask sage whether our curve has many rational points by asking it to (try to) compute the rank.


> E6.rank()
1


If the rank is at least $1$, then there are infinitely many rational points on the curve and $n$ is a congruent number. If the rank is $0$, then $n$ is not congruent.2

For the curve $Y^2 = X^3 – 36 X$ corresponding to whether $6$ is congruent, sage returns that the rank is $1$. We can ask sage to try to find a rational point on the elliptic curve through


> E6.point_search(10)
[(-3 : 9 : 1)]


The 10 in this code is a limit on the complexity of the point. The precise definition isn’t important — using $10$ is a reasonable limit for us.

We see that this output something. When sage examines the elliptic curve, it uses the equation $Y^2 Z = X^3 – 36 X Z^2$ — it turns out that in many cases, it’s easier to perform computations when every term is a polynomial of the same degree. The coordinates it’s giving us are of the form $(X : Y : Z)$, which looks a bit odd. We can ask sage to return just the XY coordinates as well.


> Pt = E6.point_search(10)[0]  # The [0] means to return the first element of the list
> Pt.xy()
(-3, 9)


In my talk, I describe a correspondence between points on elliptic curves and rational right triangles. In the talk, it arises as the choice of coordinates. But what matters for us right now is that the correspondence taking a point $(x, y)$ on an elliptic curve to a triangle $(a, b, c)$ is given by
$$(x, y) \mapsto \Big( \frac{n^2-x^2}{y}, \frac{-2 \cdot x \cdot y}{y}, \frac{n^2 + x^2}{y} \Big).$$

We can write a sage function to perform this map for us, through


> def pt_to_triangle(P):
x, y = P.xy()
return (36 - x**2)/y, (-2*x*6/y), (36+x**2)/y

> pt_to_triangle(Pt)
(3, 4, 5)


This returns the $(3, 4, 5)$ triangle!

Of course, we knew this triangle the whole time. But we can use sage to get more points. A very cool fact is that rational points on elliptic curves form a group under a sort of addition — we can add points on elliptic curves together and get more rational points. Sage is very happy to perform this addition for us, and then to see what triangle results.


> Pt2 = Pt + Pt
> Pt2.xy()
(25/4, -35/8)
> pt_to_triangle(Pt2)
(7/10, 120/7, -1201/70)


Another rational triangle with area $6$ is the $(7/10, 120/7, 1201/70)$ triangle. (You might notice that sage returned a negative hypotenuse, but it’s the absolute values that matter for the area). After scaling this to an integer triangle, we get the integer right triangle $(49, 1200, 1201)$ (and we can check that the squarefree part of the area is $6$).

Let’s do one more.


> Pt3 = Pt + Pt + Pt
> Pt3.xy()
(-1587/1369, -321057/50653)
> pt_to_triangle(Pt3)
(-4653/851, -3404/1551, -7776485/1319901)


That’s a complicated triangle! It may be fun to experiment some more — the triangles rapidly become very, very complicated. In fact, it was very important to the main result of our paper that these triangles become so complicated so quickly!

## Slides for a talk at JMM 2019

Today, I’m giving a talk on Zeroes of L-functions associated to half-integral weight modular forms, which includes some joint work with Li-Mei Lim and Tom Hulse, and which alludes to other joint work touched on previously with Jeff Hoffstein and Min Lee (and which perhaps should have been finished a few years ago).

## Update to Second Moments in the Generalized Gauss Circle Problem

Last year, my coauthors Tom Hulse, Chan Ieong Kuan, and Alex Walker posted a paper to the arXiv called “Second Moments in the Generalized Gauss Circle Problem”. I’ve briefly described its contents before.

This paper has been accepted and will appear in Forum of Mathematics: Sigma.

This is the first time I’ve submitted to the Forum of Mathematics, and I must say that this has been a very good journal experience. One interesting aspect about FoM: Sigma is that they are immediate (gold) open access, and they don’t release in issues. Instead, articles become available (for free) from them once the submission process is done. I was reviewing a publication-proof of the paper yesterday, and they appear to be very quick with regards to editing. Perhaps the paper will appear before the end of the year.

An updated version (the version from before the handling of proofs at the journal, so there will be a number of mostly aesthetic differences with the published version) of the paper will appear on the arXiv on Monday 10 December.1

## A new appendix has appeared

There is one major addition to the paper that didn’t appear in the original preprint. At one of the referee’s suggestions, Chan and I wrote an appendix. The major content of this appendix concerns a technical detail about Rankin-Selberg convolutions.

If $f$ and $g$ are weight $k$ cusp forms on $\mathrm{SL}(2, \mathbb{Z})$ with expansions $$f(z) = \sum_ {n \geq 1} a(n) e(nz), \quad g(z) = \sum_ {n \geq 1} b(n) e(nz),$$ then one can use a (real analytic) Eisenstein series $$E(s, z) = \sum_ {\gamma \in \mathrm{SL}(2, \mathbb{Z})_ \infty \backslash \mathrm{SL}(2, \mathbb{Q})} \mathrm{Im}(\gamma z)^s$$ to recognize the Rankin-Selberg $L$-function $$\label{RS} L(s, f \otimes g) := \zeta(s) \sum_ {n \geq 1} \frac{a(n)b(n)}{n^{s + k – 1}} = h(s) \langle f g y^k, E(s, z) \rangle,$$ where $h(s)$ is an easily-understandable function of $s$ and where $\langle \cdot, \cdot \rangle$ denotes the Petersson inner product.

When $f$ and $g$ are not cusp forms, or when $f$ and $g$ are modular with respect to a congruence subgroup of $\mathrm{SL}(2, \mathbb{Z})$, then there are adjustments that must be made to the typical construction of $L(s, f \otimes g)$.

When $f$ and $g$ are not cusp forms, then Zagier2 provided a way to recognize $L(s, f \otimes g)$ when $f$ and $g$ are modular on the full modular group $\mathrm{SL}(2, \mathbb{Z})$. And under certain conditions that he describes, he shows that one can still recognize $L(s, f \otimes g)$ as an inner product with an Eisenstein series as in \eqref{RS}.

In principle, his method of proof would apply for non-cuspidal forms defined on congruence subgroups, but in practice this becomes too annoying and bogged down with details to work with. Fortunately, in 2000, Gupta3 gave a different construction of $L(s, f \otimes g)$ that generalizes more readily to non-cuspidal forms on congruence subgroups. His construction is very convenient, and it shows that $L(s, f \otimes g)$ has all of the properties expected of it.

However Gupta does not show that there are certain conditions under which one can recognize $L(s, f \otimes g)$ as an inner product against an Eisenstein series.4 For this paper, we need to deal very explicitly and concretely with $L(s, \theta^2 \otimes \overline{\theta^2})$, which is formed from the modular form $\theta^2$, non-cuspidal on a congruence subgroup.

The Appendix to the paper can be thought of as an extension of Gupta’s paper: it uses Gupta’s ideas and techniques to prove a result analogous to \eqref{RS}. We then use this to get the explicit understanding necessary to tackle the Gauss Sphere problem.

There is more to this story. I’ll return to it in a later note.

## Other submission details for FoM: Sigma

I should say that there are many other revisions between the original preprint and the final one. These are mainly due to the extraordinary efforts of two Referees. One Referee was kind enough to give us approximately 10 pages of itemized suggestions and comments.

When I first opened these comments, I was a bit afraid. Having so many comments was daunting. But this Referee really took his or her time to point us in the right direction, and the resulting paper is vastly improved (and in many cases shortened, although the appendix has hidden the simplified arguments cut in length).

More broadly, the Referee acted as a sort of mentor with respect to my technical writing. I have a lot of opinions on technical writing,5 but this process changed and helped sharpen my ideas concerning good technical math writing.

I sometimes hear lots of negative aspects about peer review, but this particular pair of Referees turned the publication process into an opportunity to learn about good mathematical exposition — I didn’t expect this.

I was also surprised by the infrastructure that existed at the University of Warwick for handling a gold open access submission. As part of their open access funding, Forum of Math: Sigma has an author-pays model. Or rather, the author’s institution pays. It took essentially no time at all for Warwick to arrange the payment (about 500 pounds).

This is a not-inconsequential amount of money, but it is much less than the 1500 dollars that PLoS One uses. The comparison with PLoS One is perhaps apt. PLoS is older, and perhaps paved the way for modern gold open access journals like FoM. PLoS was started by group of established biologists and chemists, including a Nobel prize winner; FoM was started by a group of established mathematicians, including multiple Fields medalists.6

I will certainly consider Forum of Mathematics in the future.

## The wrong way to compute a sum: addendum

In my previous note, I looked at an amusing but inefficient way to compute the sum $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1}$$ using Mellin and inverse Mellin transforms. This was great fun, but the amount of work required was more intense than the more straightforward approach offered immediately by using Lambert series.

However, Adam Harper suggested that there is a nice shortcut that we can use (although coming up with this shortcut requires either a lot of familiarity with Mellin transforms or knowledge of the answer).

In the Lambert series approach, one shows quickly that $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ and then evaluates this last sum directly. For the Mellin transform approach, we might ask: do the two functions $$f(x) = \sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1}$$ and $$g(x) = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ have the same Mellin transforms? From the previous note, we know that they have the same values at $1$.

We also showed very quickly that $$\mathcal{M} [f] = \frac{1}{(\log 2)^2} \Gamma(s) \zeta(s-1).$$ The more difficult parts from the previous note arose in the evaluation of the inverse Mellin transform at $x=1$.

Let us compute the Mellin transform of $g$. We find that \begin{align} \mathcal{M}[g] &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{2^{nx}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{e^{nx \log 2}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} \frac{n}{(n \log 2)^s} \int_0^\infty x^s e^{-x} \frac{dx}{x} \notag \\ &= \frac{1}{(\log 2)^2} \zeta(s-1)\Gamma(s). \notag \end{align} To go from the second line to the third line, we did the change of variables $x \mapsto x/(n \log 2)$, yielding an integral which is precisely the definition of the Gamma function.

Thus we see that $$\mathcal{M}[g] = \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1) = \mathcal{M}[f],$$ and thus $f(x) = g(x)$. (“Nice” functions with the same “nice” Mellin transforms are also the same, exactly as with Fourier transforms).

This shows that not only is $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ but in fact $$\sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1} = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ for all $x > 1$.

I think that’s sort of slick.

## The wrong way to compute a sum

At a recent colloquium at the University of Warwick, the fact that
\label{question}
\sum_ {n \geq 1} \frac{\varphi(n)}{2^n – 1} = 2.

Although this was mentioned in passing, John Cremona asked — How do you prove that?

It almost fails a heuristic check, as one can quickly check that
\label{similar}
\sum_ {n \geq 1} \frac{n}{2^n} = 2,

which is surprisingly similar to \eqref{question}. I wish I knew more examples of pairs with a similar flavor.

[Edit: Note that an addendum to this note has been added here. In it, we see that there is a way to shortcut the “hard part” of the long computation.]

## The right way

Shortly afterwards, Adam Harper and Samir Siksek pointed out that this can be determined from Lambert series, and in fact that Hardy and Wright include a similar exercise in their book. This proof is delightful and short.

The idea is that, by expanding the denominator in power series, one has that

\sum_{n \geq 1} a(n) \frac{x^n}{1 – x^n} \notag
= \sum_ {n \geq 1} a(n) \sum_{m \geq 1} x^{mn}
= \sum_ {n \geq 1} \Big( \sum_{d \mid n} a(d) \Big) x^n,

where the inner sum is a sum over the divisors of $d$. This all converges beautifully for $\lvert x \rvert < 1$.

Applied to \eqref{question}, we find that

\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} \notag
= \sum_ {n \geq 1} \varphi(n) \frac{2^{-n}}{1 – 2^{-n}}
= \sum_ {n \geq 1} 2^{-n} \sum_{d \mid n} \varphi(d),

and as

\sum_ {d \mid n} \varphi(d) = n, \notag

we see that \eqref{question} can be rewritten as \eqref{similar} after all, and thus both evaluate to $2$.

That’s a nice derivation using a series that I hadn’t come across before. But that’s not what this short note is about. This note is about evaluating \eqref{question} in a different way, arguably the wrong way. But it’s a wrong way that works out in a nice way that at least one person1 finds appealing.

## The wrong way

We will use Mellin inversion — this is essentially Fourier inversion, but in a change of coordinates.

Let $f$ denote the function

f(x) = \frac{1}{2^x – 1}. \notag

Denote by $f^ *$ the Mellin transform of $f$,

f * (s):= \mathcal{M} [f(x)] (s) := \int_ 0^\infty f(x) x^s \frac{dx}{x}
= \frac{1}{(\log 2)^2} \Gamma(s)\zeta(s),\notag

where $\Gamma(s)$ and $\zeta(s)$ are the Gamma function and Riemann zeta functions.2

For a general nice function $g(x)$, its Mellin transform satisfies

\mathcal{M}[f(nx)] (s)
= \int_0^\infty g(nx) x^s \frac{dx}{x}
= \frac{1}{n^s} \int_0^\infty g(x) x^s \frac{dx}{x}
= \frac{1}{n^s} g^ * (s).\notag

Further, the Mellin transform is linear. Thus
\label{mellinbase}
\mathcal{M}[\sum_{n \geq 1} \varphi(n) f(nx)] (s)
= \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} f^ * (s)
= \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} \frac{\Gamma(s) \zeta(s)}{(\log 2)^s}.

The Euler phi function $\varphi(n)$ is multiplicative and nice, and its Dirichlet series can be rewritten as

\sum_{n \geq 1} \frac{\varphi(n)}{n^s} \notag
= \frac{\zeta(s-1)}{\zeta(s)}.

Thus the Mellin transform in \eqref{mellinbase} can be written as

\frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1). \notag

By the fundamental theorem of Mellin inversion (which is analogous to Fourier inversion, but again in different coordinates), the inverse Mellin transform will return the original function. The inverse Mellin transform of a function $h(s)$ is defined to be

\mathcal{M}^{-1}[h(s)] (x) \notag
:=
\frac{1}{2\pi i} \int_ {c – i \infty}^{c + i\infty} x^s h(s) ds,

where $c$ is taken so that the integral converges beautifully, and the integral is over the vertical line with real part $c$. I’ll write $(c)$ as a shorthand for the limits of integration. Thus
\label{mellininverse}
\sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1}
= \frac{1}{2\pi i} \int_ {(3)} \frac{1}{(\log 2)^s}
\Gamma(s) \zeta(s-1) x^{-s} ds.

We can now describe the end goal: evaluate \eqref{mellininverse} at $x=1$, which will recover the value of the original sum in \eqref{question}.

How can we hope to do that? The idea is to shift the line of integration arbitrarily far to the left, pick up the infinitely many residues guaranteed by Cauchy’s residue theorem, and to recognize the infinite sum as a classical series.

The integrand has residues at $s = 2, 0, -2, -4, \ldots$, coming from the zeta function ($s = 2$) and the Gamma function (all the others). Note that there aren’t poles at negative odd integers, since the zeta function has zeroes at negative even integers.

Recall, $\zeta(s)$ has residue $1$ at $s = 1$ and $\Gamma(s)$ has residue $(-1)^n/{n!}$ at $s = -n$. Then shifting the line of integration and picking up all the residues reveals that

\sum_{n \geq 1} \frac{\varphi(n)}{2^{n} – 1} \notag
=\frac{1}{\log^2 2} + \zeta(-1) + \frac{\zeta(-3)}{2!} \log^2 2 +
\frac{\zeta(-5)}{4!} \log^4 2 + \cdots

The zeta function at negative integers has a very well-known relation to the Bernoulli numbers,
\label{zeta_bern}
\zeta(-n) = – \frac{B_ {n+1}}{n+1},

where Bernoulli numbers are the coefficients in the expansion
\label{bern_gen}
\frac{t}{1 – e^{-t}} = \sum_{m \geq 0} B_m \frac{t^m}{m!}.

Many general proofs for the values of $\zeta(2n)$ use this relation and the functional equation, as well as a computation of the Bernoulli numbers themselves. Another important aspect of Bernoulli numbers that is apparent through \eqref{zeta_bern} is that $B_{2n+1} = 0$ for $n \geq 1$, lining up with the trivial zeroes of the zeta function.

Translating the zeta values into Bernoulli numbers, we find that
\eqref{question} is equal to
\begin{align}
&\frac{1}{\log^2 2} – \frac{B_2}{2} – \frac{B_4}{2! \cdot 4} \log^2 2 –
\frac{B_6}{4! \cdot 6} \log^4 2 – \frac{B_8}{6! \cdot 8} \cdots \notag \\
&=
-\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!}. \label{recog}
\end{align}
This last sum is excellent, and can be recognized.

For a general exponential generating series

F(t) = \sum_{m \geq 0} a(m) \frac{t^m}{m!},\notag

we see that

\frac{d}{dt} \frac{1}{t} F(t) \notag
=\sum_{m \geq 0} (m-1) a(m) \frac{t^{m-2}}{m!}.

Applying this to the series defining the Bernoulli numbers from \eqref{bern_gen}, we find that

\frac{d}{dt} \frac{1}{t} \frac{t}{1 – e^{-t}} \notag
=- \frac{e^{-t}}{(1 – e^{-t})^2},

and also that

\frac{d}{dt} \frac{1}{t} \frac{t}{1 – e^{-t}} \notag
=\sum_{m \geq 0} (m-1) B_m \frac{(t)^{m-2}}{m!}.

This is exactly the sum that appears in \eqref{recog}, with $t = \log 2$.

Putting this together, we find that

\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!} \notag
=\frac{e^{-\log 2}}{(1 – e^{-\log 2})^2}
= \frac{1/2}{(1/2)^2} = 2.

Thus we find that \eqref{question} really is equal to $2$, as we had sought to show.

## Notes from a Talk at Building Bridges 4

On 18 July 2018 I gave a talk at the 4th Building Bridges Automorphic Forms Workshop, which is hosted at the Renyi Institute in Budapest, Hungary this year. In this talk, I spoke about counting points on hyperboloids, with a certain focus on counting points on the three dimensional hyperboloid

$$$$X^2 + Y^2 = Z^2 + h$$$$

for any fixed integer $h$.

I gave a similar talk at the 32nd Automorphic Forms Workshop in Tufts in March. I don’t say this during my talk, but a big reason for giving these talks is to continue to inspire me to finish the corresponding paper. (There are still a couple of rough edges that need some attention).

The methodology for the result relies on the spectral expansion of half-integral weight modular forms. This is unfriendly to those unfamiliar with the subject, and particularly mysterious to students. But there is a nice connection to a topic discussed by Arpad Toth during the previous week’s associated summer school.

Arpad sketched a proof of the spectral decomposition of holomorphic modular cusp forms on $\Gamma = \mathrm{SL}(2, \mathbb{Z})$. He showed that
$$L^2(\Gamma \backslash \mathcal{H}) = \textrm{cuspidal} \oplus \textrm{Eisenstein}, \tag{1}$$
where the cuspidal contribution comes from Maass forms and the Eisenstein contribution comes from line integrals against Eisenstein series.

The typical Eisenstein series $$$$E(z, s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \textrm{Im}(\gamma z)^s$$$$ only converges for $\mathrm{Re}(s) > 1$, and the initial decomposition in $(1)$ implicitly has $s$ in this range.

To write down the integrals appearing in the Eisenstein spectrum explicitly, one normally shifts the line of integration to $1/2$. As Arpad explained, classically this produces a pole at $s = 1$ (which is the constant function).

In half-integral weight, the Eisenstein series has a pole at $s = 3/4$, with the standard theta function

$$$$\theta(z) = \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}$$$$

as the residue. (More precisely, it’s a constant times $y^{1/4} \theta(z)$, or a related theta function for $\Gamma_0(N)$). I refer to this portion of the spectrum as the residual spectrum, since it comes from often-forgotten residues of Eisenstein series. Thus the spectral decomposition for half-integral weight objects is a bit more complicated than the normal case.

When giving talks involving half-integral weight spectral expansions to audiences including non-experts, I usually omit description of this. But for those who attended the summer school, it’s possible to at least recognize where these additional terms come from.

The slides for this talk are available here.

## Paper Announcement: A Shifted Sum for the Congruent Number Problem

Tom Hulse, Chan Ieong Kuan, Alex Walker, and I have just uploaded a new paper to the arXiv titled A Shifted Sum for the Congruent Number Problem. In this charming, short paper, we investigate a particular sum of terms which are products of square-indicator functions and show that its asymptotics are deeply connected to congruent numbers. This note serves to describe and provide additional context for these results. (This note is also available as a pdf).

## Congruent Numbers

We consider some triangles. There are many right triangles, such as the triangle with sides $(3, 4, 5)$ or the triangle with sides $(1, 1, \sqrt{2})$. We call a right triangle rational when all its side lengths are rational numbers. For illustration, $(3, 4, 5)$ is rational, while $(1, 1, \sqrt{2})$ is not. $\DeclareMathOperator{\sqfree}{sqfree}$

There is mythology surrounding rational right triangles. According to legend, the ancient Greeks, led both philosophcally and mathematically by Pythagoras (who was the first person to call himself a philosopher and essentially the first to begin to distill and codify mathematics), believed all numbers and quantities were ratios of integers (rational). When a disciple of Pythagoras named Hippasus found that the side lengths of the right triangle $(1, 1, \sqrt{2})$ were not rational multiples of each other, the other followers of Pythagoras killed him by casting him overboard while at sea for having produced an element which contradicted the gods. (It with some irony that we now attribute this as a simple consequence of the Pythagorean Theorem).

This mythology is uncertain, but what is certain is that even the ancient Greeks were interested in studying rational right triangles, and they began to investigate what we now call the Congruent Number Problem. By the year 972 the CNP appears in Arabic manuscripts in (essentially) its modern formulation. The Congruent Number Problem (CNP) may be the oldest unresolved math problem.

We call a positive rational number $t$ congruent if there is a rational right triangle with area $t$. The triangle $(3,4,5)$ shows that $6 = 3 \cdot 4 / 2$ is congruent. The CNP is to describe all congruent numbers. Alternately, the CNP asks whether there is an algorithm to show definitively whether or not $t$ is a congruent number for any $t$.

We can reduce the problem to a statement about integers. If the rational number $t = p/q$ is the area of a triangle with legs $a$ and $b$, then the triangle $aq$ and $bq$ has area $tq^2 = pq$. It follows that to every rational number there is an associated squarefree integer for which either both are congruent or neither are congruent. Further, if $t$ is congruent, then $ty^2$ and $t/y^2$ are congruent for any integer $y$.

We may also restrict to integer-sided triangles if we allow ourselves to look for those triangles with squarefree area $t$. That is, if $t$ is the area of a triangle with rational sides $a/A$ and $b/B$, then $tA^2 B^2$ is the area of the triangle with integer sides $aB$ and $bA$.

It is in this form that we consider the CNP today.

Congruent Number Problem

Given a squarefree integer $t$, does there exist a triangle with integer side lengths such that the squarefree part of the area of the triangle is $t$?

We will write this description a lot, so for a triangle $T$ we introduce the notation

\sqfree(T) = \text{The squarefree part of the area of } T.

For example, the area of the triangle $T = (6, 8, 10)$ is $24 = 6 \cdot 2^2$, and so $\sqfree(T) = 6$. We should expect this, as $T$ is exactly a doubled-in-size $(3,4,5)$ triangle, which also corresponds to the congruent number $6$. Note that this allows us to only consider primitive right triangles.

## Main Result

Let $\tau(n)$ denote the square-indicator function. That is, $\tau(n)$ is $1$ if $n$ is a square, and is $0$ otherwise. Then the main result of the paper is that the sum

S_t(X) := \sum_{m = 1}^X \sum_{n = 1}^X \tau(m-n)\tau(m)\tau(nt)\tau(m+n)

is related to congruent numbers through the asymptotic

S_t(X) = C_t \sqrt X + O_t\Big( \log^{r/2} X\Big),

where

C_t = \sum_{h_i \in \mathcal{H}(t)} \frac{1}{h_i}.

Each $h_i$ is a hypotenuse of a primitive integer right triangle $T$ with $\sqfree(T) = t$. Each hypotnesue will occur in a pair of similar triangles $(a,b, h_i)$ and $(b, a, h_i)$; $\mathcal{H}(t)$ is the family of these triangles, choosing only one triangle from each similar pair. The exponent $r$ in the error term is the rank of the elliptic curve

E_t(\mathbb{Q}): y^2 = x^3 – t^2 x.

What this says is that $S_t(X)$ will have a main term if and only if $t$ is a congruent number, so that computing $S_t(X)$ for sufficiently large $X$ will show whether $t$ is congruent. (In fact, it’s easy to show that $S_t(X) \neq 0$ if and only if $t$ is congruent, so the added value here is the nature of the asymptotic).

We should be careful to note that this does not solve the CNP, since the error term depends in an inexplicit way on the desired number $t$. What this really means is that we do not have a good way of recognizing when the first nonzero term should occur in the double sum. We can only guarantee that for any $t$, understanding $S_t(X)$ for sufficiently large $X$ will allow one to understand whether $t$ is congruent or not.

## Intuition and Methodology

There are four primary components to this result:

1. There is a bijection between primitive integer right triangles $T$ with
$\sqfree(T) = t$ and arithmetic progressions of squares $m^2 – tn^2, m^2, m^2 + tn^2$ (where each term is itself a square).
2. There is a bijection between primitive integer right triangles $T$ with
$\sqfree(T) = t$ and points on the elliptic curve $E_t(\mathbb{Q}): y^2 = x^3 – t x$ with $y \neq 0$.
3. If the triangle $T$ corresponds to a point $P$ on the curve $E_t$, then
the size of the hypotenuse of $T$ can be bounded below by $H(P)$, the
(naive) height of the point on the elliptic curve.
4. Néron (and perhaps Mordell, but I’m not quite fluent in the initial
history of the theory of elliptic curves) proved strong (upper) bounds on
the number of points on an elliptic curve up to a given height. (In fact,
they proved asymptotics which are much stronger than we use).

In this paper, we use $(1)$ to relate triangles $T$ to the sum $S_t(X)$ and we use $(2)$ to relate these triangles to points on the elliptic curve. Tracking the exact nature of the hypotenuses through these bijections allows us to relate the sum to certain points on elliptic curves. In order to facilitate the tracking of these hypotenuses, we phrase these bijections in slightly different ways than have appeared in the literature. By $(3)$ and $(4)$, we can bound the number and size of the hypotenuses which appear in terms of numbers of points on the elliptic curve up to a certain height. Intuitively this is why the higher the rank of the elliptic curve (corresponding roughly to the existence of many more points on the curve), the worse the error term in our asymptotic.

I would further conjecture that the error term in our asymptotic is essentially best-possible, even though we have thrown away some information in our proof.

We are not the first to note either the bijection between triangles $T$ and arithmetic progressions of squares or between triangles $T$ and points on a particular elliptic curve. The first is surely an ancient observation, but I don’t know who first considered the relation to elliptic curves. But it’s certain that this was a fundamental aspect in Tunnell’s famous work A Classical Diophantine Problem and Modular Forms of Weight 3/2 from 1983, where he used the properties of the elliptic curve $E_t$ to relate the CNP to the Birch and Swinnerton-Dyer Conjecture.

One statement following from the Birch and Swinnerton-Dyer conjecture (BSD) is that if an elliptic curve $E$ has rank $r$, then the $L$-function $L(s, E)$ has a zero of order $r$ at $1$. The relation between lots of points on the curve and the existence of a zero is intuitive from the approximate relation that

L(1, E) \approx \lim_{X} \prod_{p \leq X} \frac{p}{\#E(\mathbb{F}_p)},

so if $E$ has lots and lots of points then we should expect the multiplicands to be very small.

On the other hand, the elliptic curve $E_t: y^2 = x^3 – t^2 x$ has the interesting property that any point with $y \neq 0$ generates a free group of points on the curve. From the bijections alluded to above, a primitive right integer triangle $T$ with $\sqfree(T) = t$ corresponds to a point on $E_t$ with $y \neq 0$, and thus guarantees that there are lots of points on the curve. Tunnell showed that what I described as “lots of points” is actually enough points that $L(1, E)$ must be zero (assuming the relation between the rank of the curve and the value of $L(1, E)$ from BSD).

Tunnell proved that if BSD is true, then $L(1, E) = 0$ if and only if $n$ is a congruent number.

Yet for any elliptic curve we know how to compute $L(1, E)$ to guaranteed accuracy (for instance by using Dokchitser’s algorithm). Thus a corollary of Tunnell’s theorem is that BSD implies that there is an algorithm which can be used to determine definitively whether or not any particular integer $t$ is congruent.

This is the state of the art on the congruent number problem. Unfortunately, BSD (or even the somewhat weaker between BSD and mere nonzero rank of elliptic curves as is necessary for Tunnell’s result for the CNP) is quite far from being proven.

In this context, the main result of this paper is not as effective at actually determining whether a number is congruent or not. But it does have the benefit of not relying on any unknown conjecture.

And there is some potential follow-up questions. The sum $S_t(X)$ appears as an integral transform of the multiple Dirichlet series

\sum_{m,n} \frac{\tau(m-n)\tau(m)\tau(nt)\tau(m+n)}{m^s n^w}
\approx
\sum_{m,n} \frac{r_1(m-n)r_1(m)r_1(nt)r_1(m+n)}{m^s n^w},

where $r_1(n)$ is $1$ if $n = 0$ or $2$ if $n$ is a positive square, and $0$ otherwise. Then $r_1(n)$ appears as the Fourier coefficients of the half-integral weight standard theta function

\theta(z)
= \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}
= \sum_{n \geq 0} r_1(n) e^{2 \pi i n z},

and $S_t(X)$ is a shifted convolution sum coming from some products of modular forms related to $\theta(z)$.

It may be possible to gain further understanding of the behavior of $S_t(X)$ (and therefore the congruent number problem) by studying the shifted convolution as coming from theta functions.

I would guess that there is a deep relation to Tunnell’s analysis in his 1983 paper, as in some sense he constructs appropriate products of three theta functions and uses them centrally in his proof. But I do not understand this relationship well enough yet to know whether it is possible to deepen our understanding of the CNP, BSD, or Tunnell’s proof. That is something to explore in the future.

## Notes from a talk at Tufts, Automorphic Forms Workshop

On 19 March I gave a talk at the 32nd Automorphic Forms Workshop, which ishosted by Tufts this year. The content of the talk concerned counting points on hyperboloids, and inparticular counting points on the three dimensional hyperboloid

$$X^2 + Y^2 = Z^2 + h$$

for any fixed integer $h$. But thematically, I wanted to give another concrete example of using modularforms to compute some sort of arithmetic data, and to mention how the perhapsapparently unrelated topic of spectral theory appears even in such an arithmeticapplication.

Somehow, starting from counting points on $X^2 + Y^2 = Z^2 + h$ (which appearssimple enough on its own that I could probably put this in front of anelementary number theory class and they would feel comfortable experimentingaway on the topic), one gets to very scary-looking expressions like

$$\sum_{t_j} \langle P_h^k, \mu_j \rangle \langle \theta^2 \overline{\theta} y^{3/4}, \mu_j \rangle + \sum_{\mathfrak{a}}\int_{(1/2)} \langle P_h^k, E_h^k(\cdot, u) \rangle \langle \theta^2 \overline{\theta} y^{3/4}, E_h^k(\cdot, u) \rangle du,$$

which is full of lots of non-obvious symbols and is generically intimidating.

Part of the theme of this talk is to give a very direct idea of how one gets tothe very complicated spectral expansion from the original lattice-countingproblem. Stated differently, perhaps part of the theme is to describe a simple-lookingnail and a scary-looking hammer, and show that the hammer actually works quitewell in this case.

The slides for this talk are available here.

## Slides from a talk at the Joint Math Meetings 2018

I’m in San Diego, and it’s charming here. (It’s certainly much nicer outside than the feet of snow in Boston. I’ve apparently brought some British rain with me, though).

Today I give a talk on counting lattice points on one-sheeted hyperboloids. These are the shapes described by
$$X_1^2 + \cdots + X_{d-1}^2 = X_d^2 + h,$$
where $h > 0$ is a positive integer. The question is: how many lattice points $x$ are on such a hyperboloid with $| x |^2 \leq R$; or equivalently, how many lattice points are on such a hyperboloid and contained within a ball of radius $\sqrt R$ centered at the origin?

I describe my general approach of transforming this into a question about the behavior of modular forms, and then using spectral techniques from the theory of modular forms to understand this behavior. This becomes a question of understanding the shifted convolution Dirichlet series
$$\sum_{n \geq 0} \frac{r_{d-1}(n+h)r_1(n)}{(2n + h)^s}.$$
Ultimately this comes from the modular form $\theta^{d-1}(z) \overline{\theta(z)}$, where
$$\theta(z) = \sum_{m \in \mathbb{Z}} e^{2 \pi i m^2 z}.$$

Here are the slides for this talk. Note that this talk is based on chapter 5 of my thesis, and (hopefully) soon a preprint of this chapter ready for submission will appear on the arXiv.

## Introduction

The primary purpose of this note is to collect a few hitherto unnoticed or unpublished results concerning gaps between powers of consecutive primes. The study of gaps between primes has attracted many mathematicians and led to many deep realizations in number theory. The literature is full of conjectures, both open and closed, concerning the nature of primes.

In a series of stunning developments, Zhang, Maynard, and Tao12 made the first major progress towards proving the prime $k$-tuple conjecture, and successfully proved the existence of infinitely many pairs of primes differing by a fixed number. As of now, the best known result is due to the massive collaborative Polymath8 project,3 which showed that there are infinitely many pairs of primes of the form $p, p+246$. In the excellent expository article, 4 Granville describes the history and ideas leading to this breakthrough, and also discusses some of the potential impact of the results. This note should be thought of as a few more results following from the ideas of Zhang, Maynard, Tao, and the Polymath8 project.

Throughout, $p_n$ will refer to the $n$th prime number. In a paper, 5 Andrica conjectured that
\label{eq:Andrica_conj}
\sqrt{p_{n+1}} – \sqrt{p_n} < 1

holds for all $n$. This conjecture, and related statements, is described in Guy’s Unsolved Problems in Number Theory.
6 It is quickly checked that this holds for primes up to $4.26 \cdot 10^{8}$ in sagemath

# Sage version 8.0.rc1
# started with sage -ipython

# sage has pari/GP, which can generate primes super quickly
from sage.all import primes_first_n

# import izip since we'll be zipping a huge list, and sage uses python2 which has
# non-iterable zip by default
from itertools import izip

# The magic number 23150000 appears because pari/GP can't compute
# primes above 436273290 due to fixed precision arithmetic
ps = primes_first_n(23150000)    # This is every prime up to 436006979

# Verify Andrica's Conjecture for all prime pairs = up to 436006979
gap = 0
for a,b in izip(ps[:-1], ps[1:]):
if b**.5 - a**.5 > gap:
A, B, gap = a, b, b**.5 - a**.5
print(gap)
print("")
print(A)
print(B)


In approximately 20 seconds on my machine (so it would not be harder to go much higher, except that I would have to go beyond pari/GP to generate primes), this completes and prints out the following output.

0.317837245196
0.504017169931
0.670873479291

7
11


Thus the largest value of $\sqrt{p_{n+1}} – \sqrt{p_n}$ was merely $0.670\ldots$, and occurred on the gap between $7$ and $11$.

So it appears very likely that the conjecture is true. However it is also likely that new, novel ideas are necessary before the conjecture is decided.

Andrica’s Conjecture can also be stated in terms of prime gaps. Let $g_n = p_{n+1} – p_n$ be the gap between the $n$th prime and the $(n+1)$st prime. Then Andrica’s Conjecture is equivalent to the claim that $g_n < 2 \sqrt{p_n} + 1$. In this direction, the best known result is due to Baker, Harman, and Pintz, 7 who show that $g_n \ll p_n^{0.525}$.

In 1985, Sandor 8 proved that $$\label{eq:Sandor} \liminf_{n \to \infty} \sqrt[4]{p_n} (\sqrt{p_{n+1}} – \sqrt{p_n}) = 0.$$ The close relation to Andrica’s Conjecture \eqref{eq:Andrica_conj} is clear. The first result of this note is to strengthen this result.

Theorem

Let $\alpha, \beta \geq 0$, and $\alpha + \beta < 1$. Then
\label{eq:main}
\liminf_{n \to \infty} p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) = 0.

We prove this theorem below. Choosing $\alpha = \frac{1}{2}, \beta = \frac{1}{4}$ verifies Sandor’s result \eqref{eq:Sandor}. But choosing $\alpha = \frac{1}{2}, \beta = \frac{1}{2} – \epsilon$ for a small $\epsilon > 0$ gives stronger results.

This theorem leads naturally to the following conjecture.

Conjecture

For any $0 \leq \alpha < 1$, there exists a constant $C(\alpha)$ such that

p_{n+1}^\alpha – p_{n}^\alpha \leq C(\alpha)

for all $n$.

A simple heuristic argument, given in the last section below, shows that this Conjecture follows from Cramer’s Conjecture.

It is interesting to note that there are generalizations of Andrica’s Conjecture. One can ask what the smallest $\gamma$ is such that

p_{n+1}^{\gamma} – p_n^{\gamma} = 1

has a solution. This is known as the Smarandache Conjecture, and it is believed that the smallest such $\gamma$ is approximately

\gamma \approx 0.5671481302539\ldots

The digits of this constant, sometimes called “the Smarandache constant,” are the contents of sequence A038458 on the OEIS. It is possible to generalize this question as well.

Open Question

For any fixed constant $C$, what is the smallest $\alpha = \alpha(C)$ such that

p_{n+1}^\alpha – p_n^\alpha = C

has solutions? In particular, how does $\alpha(C)$ behave as a function of $C$?

This question does not seem to have been approached in any sort of generality, aside from the case when $C = 1$.

## Proof of Theorem

The idea of the proof is very straightforward. We estimate \eqref{eq:main} across prime pairs $p, p+246$, relying on the recent proof from Polymath8 that infinitely many such primes exist.

Fix $\alpha, \beta \geq 0$ with $\alpha + \beta < 1$. Applying the mean value theorem of calculus on the function $x \mapsto x^\alpha$ shows that
\begin{align}
p^\beta \big( (p+246)^\alpha – p^\alpha \big) &= p^\beta \cdot 246 \alpha q^{\alpha – 1} \\\
&\leq p^\beta \cdot 246 \alpha p^{\alpha – 1} = 246 \alpha p^{\alpha + \beta – 1}, \label{eq:bound}
\end{align}
for some $q \in [p, p+246]$. Passing to the inequality in the second line is done by realizing that $q^{\alpha – 1}$ is a decreasing function in $q$. As $\alpha + \beta – 1 < 0$, as $p \to \infty$ we see that\eqref{eq:bound} goes to zero.

Therefore

\liminf_{n \to \infty} p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) = 0,

as was to be proved.

## Further Heuristics

Cramer’s Conjecture states that there exists a constant $C$ such that for all sufficiently large $n$,

p_{n+1} – p_n < C(\log n)^2.

Thus for a sufficiently large prime $p$, the subsequent prime is at most $p + C (\log p)^2$. Performing a similar estimation as above shows that

(p + C (\log p)^2)^\alpha – p^\alpha \leq C (\log p)^2 \alpha p^{\alpha – 1} =
C \alpha \frac{(\log p)^2}{p^{1 – \alpha}}.

As the right hand side vanishes as $p \to \infty$, we see that it is natural to expect that the main Conjecture above is true. More generally, we should expect the following, stronger conjecture.

Conjecture’

For any $\alpha, \beta \geq 0$ with $\alpha + \beta < 1$, there exists a constant $C(\alpha, \beta)$ such that

p_n^\beta (p_{n+1}^\alpha – p_n^\alpha) \leq C(\alpha, \beta).