From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: triangles with integer sides and area - seeking guide to literature
Date: 13 Sep 1996 18:05:19 GMT
Keywords: triangle integer guide
In article ,
Mike Ruxton (CHS) wrote:
>My interest has recently been piqued by the problem of for which
>integers N there exists a triangle with integer sides with area equal
>to N. The original problem was restricted - I believe - to right
>triangles. I believe the problem reduces to solving elliptic curve
>equations.
If a triangle has sides of lengths a b and c, then the area N
may be determined from Heron's formula,
N^2 = s(s-a)(s-b)(s-c)
where s = (a+b+c)/2. Thus your question may be rephrased as, "For which
values of N does the equation
16N^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)
have solutions in positive integers a, b, and c?".
For a specific N, this question may be resolved simply by checking
all possible factorizations of (4N)^2 into four positive factors
to see if the largest factor is the sum of the other three.
The right side of this equation equals
2(a^2b^2 + a^2c^2 + b^2c^2) - (a^4+b^4+c^4).
You may let x = b/a, y = c/a, and r=4N/a^2; then the problem essentially
asks, for what rational numbers r is there a rational point on the curve
x^4 + y^4 + r^2 = 2(x^2 y^2 + x^2 + y^2) ?
I don't know if there is an easy answer to this, but yes, I suppose this is a
question about elliptic curves. I don't know what literature is
available.
I would guess though that you heard about a different problem when
right triangles were mentioned. It's pretty easy to determine for each
integer N whether it's the are of an _integral_ right triangle
(especially if you know how to factor N). The more interesting
(or at least, harder) question is whether there is a _rational_ right
triangle with area N. This question was evidently of interest to the
ancients. It's equivalent to asking if there is an arithmetic progression
of three (rational) squares with common difference N.
Now it is not clear that this question can be resolved after a finite
amount of computation. For example, when N=5 there is the triangle with
sides 3/2, 20/3, and 41/6 which is a right triangle with rational
sides and area N. But how can one know which denominators to expect?
It was not until Euler's day that a solution for N=7 was found
(a=24/5, b=35/12) and the proof that there is no solution for N=1 is
equivalent to Fermat's Last Theorem with exponent 4 (which _was_ proved
by Fermat).
Suppose we have a right triangle with legs of length a and b and
a hypotenuse of length r. Suppose the area of the triangle is N.
Then
N = (1/2) ab
r^2 = a^2 + b^2.
Rational solutions to the last equation are well known, so let me
summarize by noting that if rational numbers a, b, r, and N
satisfy these two equations, then
x = -Na/(b+r) and y = 2N^2/(b+r)
are also rational and satisfy the equation
y^2 = x^3 - N^2 x;
conversely, given a rational solution to this single equation we may set
a = -2Nx/y b = (N^2-x^2)/y and r = (N^2 + x^2)/y,
three rational numbers which satisfy the original pair of equations.
(Assuming y <> 0, of course.) Checking quickly the signs of possible
solutions we see
Theorem: There is a rational right triangle with area N iff there is a
rational point on the curve y^2 = x^3 + D x (where D = -N^2) other
than the points with y=0.
Actually it is known that, except for the case D=4, the torsion points
on this curve are precisely the points with y=0, so the theorem may
just as well be written to say "... iff the above curve has rank at
least 1 over the rationals." That, in turn, shows that if there are
any rational right triangles with area N, then there are infinitely many!
(For example, when N=5, from the first solution above we can find a second
with x = (41/12)^2, which leads to the right triangle with area 5 and
legs of length 4920/1519 and 1519/42.)
The theorem above is an interesting one, but it is made less useful by the
fact that determining the rank of an elliptic curve is at best indirect
and (should some likely conjectures turn out to fail) possibly
uncomputable with existing algorithms. For example, some experimentation
might convince you that there is no rational point on the curve
y^2 = x^3 + 877 x.
But this would simply be a spectacular failure of the
"law of small numbers": your experiments would have led you astray,
since there _are_ solutions but the simplest one has
x = (612776083187947368101 / 7884153586063900210) ^ 2
in lowest terms!
The most concrete result in this direction, as far as I know, is
due to Tunnell. Given an integer N let n be its square-free part.
Let c1 be the number of triples of integers satsifying
2x^2 + y^2 + 8z^2 = n
and let c2 be the number of triples of integers satisfying
2x^2 + y^2 + 32z^2 = n.
Then
(1) if N is the area of a right triangle with rational sides,
then c1 = 2 c2.
(2) if c1 = 2 c2 and the Birch--Swinnerton-Dyer conjecture is
true, then N is the area of a rational right triangle.
There is a great body of evidence supporting the BSD conjecture, and
the full weight of it is not needed for the proof of this theorem, so
I would say most people would credit Tunnell's result with characterizing
the set of values of N.
Note that this theorem does not show how to _find_ the triangle once
it exists. For some small integers N even the simplest solutions are
delightfully complicated (e.g. N=157).
Integers N for which the elliptic curve above has a nontrivial rational point
are called _congruent numbers_ (terminology which I find to be needlessly
confusing, but now well established). This topic is taken as the
introduction and point of reference in Neal Koblitz's book on Elliptic
Curves, where most of these details may be found.
dave
==============================================================================
From: wcw@math.psu.edu (William C Waterhouse)
Newsgroups: sci.math
Subject: triangles with integer sides and area
Date: 13 Sep 1996 21:52:35 GMT
Organization: Dept. of Mathematics, Penn State
A question on this topic was posted about a week ago, and I've
now managed to track down my vague memory of it.
According to Math Reviews (MR 31, #121), the problem was solved in
Popovici, Constantin P., Heronian triangles [in Russian],
Revue [roumaine] de mathematiques pures at appliquees,
7 (1962), 439-457.
The stated solution is (in essence) that all solutions are
given by sides
a = ((t*e*r)^2 + u*x*s)^2)*v*z,
b = (u*v*s^2 - t*z*r^2)*(u*z*x^2 + t*v*e^2),
c = ((z*x*r)^2 + (v*e*s)^2)*u*t,
where the variables are natural numbers and the first factor in b is
positive.
More precisely, if f is the greatest common divisor of the
two factors in b, then f divides all three expressions, and the most
general expression is found by dividing by f and multiplying by
a random natural number.
It's easy to use a program like Mathematica to verify that the
area for the a,b,c given above is indeed an integer, namely
Area = e*r*s*t*u*v*x*z*b.
There is no sketch of proof in the review, and I have not read
the article myself.
William C. Waterhouse
Penn State
==============================================================================
From: Rusin
To: Rusin
Addenda (notes from Koblitz' book)
non-congruent numbers are 1 (Fermat - n=4 case of FLT), 2, 3, 4, ...
congruent numbers are 5, 6, 7 (Euler), ...
Tunnel: for n square free we consider several assertions;
a. n is congruent
b. Elliptic curve has rank > 0
c. L(E,1) = 0 (the L-function of E vanishes at 1)
d. certain coefficients in theta functions vanish
e. #{(x,y,z): 2x^2+y^2+8z^2=n}=2*#{(x,y,z): 2x^2+y^2+32z^2=n}
Then
a <=> b. Coates-Wiles: b => c; BSD conjecture: c => b.
Tunnel: c <=> d <=> e.
(e) holds if n = 5 or 7 mod 8.
Also, (a) iff:
there is a 3-term arithmetic progression of rational perfect
squares and common difference n.
hard to find, e.g. n=157 => x = a/b:
a=411340319227716149383203 (2nd 3 may be a 5?)
b= 21666555693714761309610