A couple of years ago, I wrote a paper *Congruent number triangles with the
same hypotenuse* with an appendix by Brendan
Hassett. This paper details a computational experiment concerning triangles,
related to the main term in my earlier paper *A shifted sum for the congruent
number problem* from 2018.

For no deep reason, I hadn't given either of these papers a dedicated discussion page here. But to my surprise, I've received many emails about them asking some questions.

This post serves as the discussion page for both papers. Below, I briefly describe what these two papers prove and how they relate to each other.

If I get further questions or comments, I'll add them to this post.

## A shifted sum for the congruent number problem

With my frequent collaborators Thomas Hulse, Chan Ieong Kuan, and Alexander
Walker, I wrote a paper describing an
explicit relationship between a certain shifted convolution sum and asymptotics
for congruent numbers.^{1}
^{1}Written in 2018, published in the *Ramanujan
Journal* in 2020.

We call a number **congruent** if it appears as the area of a right triangle
with rational sidelengths. For example, the triangle with sidelengths $(3, 4,
5)$ is a right triangle with area $6$, and thus $6$ is a congruent number.
The decision problem for congruent numbers is famously hard.^{2}
^{2}Here, we mean
an algorithm that is guaranteed to terminate. Thus "try all triangles" isn't an
acceptable algorithm.

QuestionGiven $q \in \mathbb{Q}$, does there exist an algorithm that determines if $q$ is congruent?

Scaling a rational right triangle with area $A$ by a rational $\lambda$ gives another rational right triangle with area $\lambda^2 A$. For example, we see that $3/2 = 6 \cdot 2^{-2}$ is congruent because we can scale the $(3, 4, 5)$ triangle by $2^{-1}$, giving $(3/2, 2, 5/2)$. Thus, after scaling, we lose no generality by only considering whether a given integer $t$ is congruent. (That is, nothing is lost in thinking only of integral areas instead of any potential rational area).

Further, scaling allows us to consider only squarefree integer areas $t$, as we can scale away the square factor. This leads to the following (equivalent) decision problem.

Equivalent QuestionGiven squarefree $t \in \mathbb{Z}$, does there exist an algorithm that determines if $t$ is congruent?

Thought of differently, the triangles $\lambda (3, 4, 5)$ all correspond to
the "same" triangle and congruent number. Given any right triangle $(a, b, c)$
with rational sidelengths, there is a unique $\lambda$ such that $(\lambda a,
\lambda b, \lambda c)$ are all integers and $\gcd(\lambda a, \lambda b, \lambda
c) = 1$. We think of this right triangle as the **primitive** right triangle,
and it is how we think about the whole class of triangle $(\lambda a, \lambda
b, \lambda c)$. The area of a primitive right triangle will be a squarefree
integer, and in fact these two descriptions ultimately offer two descriptions
of the same class of fundamental elements.

There is a famous conjectural resolution due to
Tunnell, which asserts that
the answer is YES *if we assume the Birch and Swinnerton-Dyer Conjecture*.

Let $\tau(n) = 1$ if $n$ is a square and $0$ if $n$ is not a square.
The primary theorem of our paper *A shifted sum for the Congruent number
problem* is the following.

Let $r$ be the rank of the elliptic curve $E_t: Y^2 = X^3 - t^2 X$ over $\mathbb{Q}$. For $X > 1$, define the shifted partial sum \begin{equation*} S_t(X) = \sum_{m = 1}^X \sum_{n = 1}^X \tau(m + n) \tau(m - n) \tau(m) \tau(tn). \end{equation*} Then \begin{equation*} S_t(X) = C_t \sqrt{X} + O_t\big( (\log X)^{\frac{r}{2}} \big), \end{equation*} where \begin{equation*} C_t = \sum \frac{1}{h_i}, \end{equation*} where the sum is over the set of hypotenuses $h_i$ or dissimilar primitive right triangles with squarefree part of the area equal to $t$.

In particular, $C_t \neq 0$ if and only if $t$ is a squarefree congruent number. If we could bound $C_t$ from below when it was nonzero, and if we could bound the implicit constant in the $O_t(\cdot)$ error term, then we would solve the congruent number decision problem.

But we don't, and these are very hard. One can show that the set of hypotenuses
of primitive triangles corresponding to a given congruent $t$ are
logarithmically sparse (i.e. there are only powers of $\log(X)$ many up to size
$X$). This means that the sum $C_t$ converges *extremely* quickly and
is morally approximately equal to the first couple of summands. Thus
approximating $C_t$ amounts to estimating the smallest hypotenuse of a
potential congruent triangle for $t$.

There is a well-known correspondence between congruent number triangles $(a, b, c)$ with area $t$ and three term arithmetic progressions of rational squares with common difference $t$. Indeed, this paper was a warmup for our later paper on counting three term arithmetic progressions, which uses similar technology. There is also a well-known correspondence between three-term arithmetic progressions of rational squares with common difference $t$ and rational points on the elliptic curve $E_t$.

The whole game of this paper is tracking through these correspondences closely. In particular, the term \begin{equation*} \tau(m + n) \tau(m - n) \tau(m) \tau(tn) \end{equation*} detects three term arithmetic progressions of rational squares with common difference $t$.

The smallest hypotenuse $h_i$ or a primitive integer right triangle showing that the squarefree integer $t$ is congruent is then very closely related to upper bounds on the height of the first rational point on the elliptic curve $E_t$. This is hard to bound!

I suspect that with a lot of work, it actually *would be possible* to handle
the implicit constant (depending on $t$) in the error term $O_t(\cdot)$. But
without being able to estimate the complexity of the smallest rational points
on elliptic curves, this wouldn't suffice.

### Aside

We had an idea for how to approach the congruent number problem through this perspective, studying the natural multiple Dirichlet series

\begin{equation*} D(s, w) = \sum_{m, n \geq 1} \frac{\tau(m + n) \tau(m - n) \tau(m) \tau(nt)}{m^s n^w}. \end{equation*}

Of note is that $D(s, s)$ would have a pole at $s = 1/2$ if and only if $t$ was congruent. So the decision problem would reduce to making sense of this multiple Dirichlet series. Presumably, having a pole would reduce precisely to behavior of a particular elliptic curve. In fact, Tunnell proved his conjectured algorithm for congruent number detection by constructing particular products of four theta functions and studying their behavior; the construction of $D(s, w)$ would also involve products of four theta functions. Perhaps they're deeply connected in ways I don't understand.

Actually making sense of this proved very hard. We don't know how to proceed and are not actively pursuing this. Our paper on arithmetic progressions can be thought of as a successful stepping stone in that direction though.

## Congruent Number Triangles with the Same Hypotenuse

I wrote *Congruent number triangles with the same hypotenuse*
later.^{3}
^{3}Written in 2020. Published in 2021 in *Simon's Symposia*.

The main question had to do with a potential complication in applying the shifted-sum asymptotic above to actually determine if a number is congruent.

Heuristically, it seems that $C_t$ is very well-approximated by exactly the reciprocal of the smallest hypotenuse. (This refers to the previous paper, described above). All other hypotenuses of relevant primitive congruent triangles are much, much smaller.

It would be a bit complicated if there were several hypotenuses all
approximately as small as the smallest. For deep reasons, there can't be more
than a fixed small multiple of the rank of the corresponding elliptic curve
$E_t$. In the worst case, there would be several *different* triangles that all
had the *exact same* hypotenuse.

In this new paper, we set out to investigate if this worst case can ever happen.

Question: can two dissimilar primitive right integer triangles have the same hypotenuse and the same squarefree part of their area?

I conjecture that this cannot happen. In an appendix to our paper, Brendan Hassett explains geometrically why this is probably an intractable problem to resolve. This has to do with the variety defined by the equations \begin{align*} s^2 + t^2 &= S^2 + T^2 \\ u^2 st(s^2 - t^2) &= v^2 ST (S^2 - T^2), \end{align*} considered in $\mathbb{P}^3_{[s, t, S, T]} \times \mathbb{P}^1_{[u, v]}$. (The point is that this admits a resolution of singularities that is of general type and simply connected. It's too complicated!).

Observe that each adjective is essential. The triangles $(3, 4, 5)$ and $(4, 3, 5)$ obviously have the same hypotenuse and area, but are similar.

Primitivity also matters. Otherwise, one could choose any two dissimilar
triangles with the same squarefree part of the area and scale them to produce a
counterexample.^{4}
^{4}And in fact if there is one triangle, there are actually
infinitely many dissimilar triangles with the same squarefree part of the area.
They are logarithmically dense.
For example, the triangles $(20, 21,
29)$ and $(12, 35, 37)$ both have area $210$, but different hypotenuseses.
However we can scale the first by $37$ and and second by $29$ to get the
triangles $(740, 777, 1073)$ and $(348, 1015, 1073)$. These have the same
hypotenuse. The first has area $210 \cdot 37^2$ and the second has area $210
\cdot 29^2$, and thus correspond to the same congruent number. But neither is
primitive, and so this doesn't give a counterexample.

(This gives an infinite family of near-counterexamples in a sense, but none of them work).

The rest of the paper goes towards an efficient experimental investigation of
this problem. I investigate two potential sources of problems: small congruent
numbers (the law of small numbers suggests that distributions of small numbers
should have more coincidental overlaps; this is even more true given that
everything is only logarithmically dense here) and congruent number elliptic
curves with high rank. Thus I investigated all congruent numbers up to $1000$
and every known congruent number elliptic curve found of rank $\geq 6$.^{5}
^{5} I
have separately and since done the same for all congruent numbers up to $2000$
and every congruent number elliptic curve of rank $4$ and $5$.

From a purely statistical standpoint, if the first several hypotenuses don't
collide, then we shouldn't expect any to collide. The points are too sparse for
random collisions. I limited my experiments by size: I considered all triangles
whose side-lengths took less than one megabyte each to store (approximately).
That is, I considered all triangles with sidelengths up to $2^{1000000}$,
approximately.^{6}
^{6}But to reiterate, the triangles themselves are
logarithmically dense, i.e. they grow exponentially. This doesn't actually lead
to *that many* triangles. It does lead to some *very large* triangles
though.

### Leave a comment

## Info on how to comment

To make a comment, please send an email using the button below. Your email
address **won't be shared** (unless you include it in the body
of your comment). If you don't want your real name to be used next to your
comment, please specify the name you would like to use. If you want your name
to link to a particular url, include that as well.

**bold**, *italics*, and plain text are allowed in
comments. A reasonable subset of markdown is supported, including lists,
links, and fenced code blocks. In addition, math can be formatted using
`$(inline math)$`

or `$$(your display equation)$$`

.

**Please use plaintext email** when commenting. See
Plaintext Email and
Comments on this site for more. Note also that
**comments are expected to be open, considerate, and
respectful.**

Comments (5)2023-11-01 RBDavid, the triangles $(740, 777, 1073)$ and $(348, 1015, 1073)$ both have the same hypotenuse. They both have the same non-square area $210$. In your paper you say these triangles do not exist? Did I miss something?

2023-11-01 DLDThank you for your comment RB. The conjecture refers to

primitivetriangles for exactly this reason. The description above now includes your example in the description.2023-11-01 RBAh, I missed "primitive" so I will go back to my GP-Pari routines, since I have an efficient way of finding square pairs summing to a third square.

2023-11-01 DLDGood luck! I'll note that the sparseness means that it is

extremely unlikelythat a naive search, even very efficiently coded, would beat the approach guided by elliptic curves I describe in the paper. But I would be very interested to know a counterexample (if it exists) or a proof (if it exists) to my conjecture.2024-02-15 PJ@RB can't find more. You already found them all