What does it really mean to “use” an idea–any idea?

I worry that I’ve totally missed the point of your question. If so, please do let me know.

Thanks for any insight!

I am working on writing a paper on the congruent number problem for a modular forms class. The goal is not that extravagant, I need to exhibit 5 is congruent using the Shimura lift, Tunnel’s Theaorem, Eichler Shimura, and Waldspurger. However, I am having a bit of trouble with the construction. Is there a nuts and bolts reference to the lift with examples available anywhere. Thank you if you can help me. Please email me if you can because I happened upon this forum only and may not find it again. Thank you so much.

Steve Winburn
secuervo@aol.com
winburn@math.uga.edu

n is even and the equation n = 8x^2 + 2y^2 + 16z^2 has twice as many integer solutions as the equation n = 8x^2 + 2y^2 + 64z^2; or

n is odd and the equation n = 2x^2 + y^2 + 8z^2 has twice as many integer solutions as the equation n = 2x^2 + y^2 + 32z^2.

This is, on the face of it, a seemingly arbitrary, unmotivated definition. One might equally arbitrarily define, say a “Hayes number” to be an N for which the equation n = x^2 + y^2 has twice as many integer solutions as the equation n = xy. But the motivation for Tunnell numbers is provided by Tunnell’s great theorem, which can be stated in two parts, as follows:

1. Every congruent number is a Tunnell number; and
2. If the BSD conjecture is true, then every Tunnell number is a congruent number.

The current state of affairs can be summarized thusly:

1. Every number (square-free or not) in the residue classes 5, 6, and 7 mod 8 is a Tunnell number; and
2. Up to a trillion, the residue classes 1, 2, and 3 mod 8 contain a total of 3,148,379,694 square-free Tunnell numbers.

If and when the BSD conjecture becomes a theorem, the distinction between Tunnell numbers and congruent numbers will evaporate. Until then, there will be additional results of the form “all Tunnell numbers up to x are congruent numbers” and “all Tunnell numbers that [blah blah blah] are congruent numbers.” A result of the first form can be obtained in a seemingly straightforward way by finding a right rational triangle of area N for each Tunnell number N up to whatever x you have the resources to compute — except that absent the BSD conjecture, there’s no algorithm that’s guaranteed to terminate (and even if you assume a guarantee of termination, some of the rational numbers can have huge numerators and denominators). I’m not sure what the current largest known x is, but I think it’s only in the thousands.

It’s worth noting that even if the BSD conjecture turns out to be false, the first part of Tunnell’s theorem is still an amazingly powerful result. It provides a very easy way to show that certains numbers cannot be the area of any right rational triangle. Prior to Tunnell’s breakthrough, this was, if anything, the harder thing to prove, since it’s often easier to prove things do exist (e.g., by finding a right rational triangle of given area) than to prove they don’t (e.g., there’s no n-th power of any positive integer that’s the sum of two other n-th powers of positive integers).

I hope this helps.

Good article.

Actually, the “both” is incorrect.

The triple generated by Euclid’s formula is primitive if and only if m and n are coprime and exactly one of them is even. (http://en.wikipedia.org/wiki/Pythagorean_triple)

You can derive this by substituting (2n) & (2p) for a & b in a^2 + b^2 = c^2. You would get c^2 is even always which is not accurate.

But also, it implies that the given number represents the count of *all* square-free congruent numbers up to a trillion. As I understand it (and I hope someone will jump in and correct me if I’m wrong), it only counts the square-free (candidate) congruent numbers up to a trillion that are congruent, modulo 8, to 1, 2, or 3.

This is because these are the only cases for which an actual computation is required to check if N satisfies the Tunnell criterion. The square-free numbers congruent to 5, 6, and 7 mod 8 automatically satisfy the criterion for the simple reason that the ternary quadratic equations N = 2x^2 + y^2 + 32z^2 and N = 2x^2 + y^2 + 8z^2 (for N odd) and N = 8x^2 + 2y^2 + 64z^2 and N = 8x^2 + 2y^2 + 16z^2 (for N even) have no integer solutions whatsoever when N=5, 6, or 7 mod 8, for bordline trivial reasons based on the observation that, mod 8, the only odd square is 1 and the only even squares are 0 and 4. When you mod out by 8, the quadratic equations reduce to N = 2x^2 + y^2 for odd N and N = 2y^2 for even N, and it’s easy to see that neither of these has any solutions mod 8 when N=5, 6, or 7 mod 8. (For example, 2y^2 can equal 0 or 2 mod 8, but never 6.) This, by the way, helps account for the curiosity you observed in your graphic.

There are 125 billion numbers in each residue class mod 8 (up to a trillion), and about 90% of them are square-free in the classes congruent to 1, 2, 3, 5, 6, and 7. (It’s reasonably straightforward to count the exact number of square-free numbers in each residue class, but I don’t trust myself to do the calculations correctly. There are, of course, *no* square-free numbers in the classes congruent to 0 or 4 mod 8.) Thus there are well over 300 billion square-free (candidate) congruent numbers altogether in the residue classes 5, 6, and 7 mod 8, whereas the recent computation showed that the residue classes 1, 2, and 3 contain only a little over 3 billion such numbers.

On the one hand, it would be nice to know the complete count of (candidate) congruent numbers, square-free or not, up to various numbers x — a kind of analog to the prime number function pi(x). On the other hand, what the recent computation was really aimed at, as I understand it, is exploring implications of the BSD conjecture for elliptic curves (e.g., the asymptotic formula mentioned by Rubinstein), and for those purposes only the square-free numbers in the residue classes 1, 2, and 3 are of interest.

See http://www.warwick.ac.uk/~masfaw/congruent.pdf for the details.