Impossible Heronian Triangles (Ratio of 2 Sides)

Question

There is no Heronian triangle (or simply consider triangles on an integer lattice which also have integer side lengths) for which one side is half the length of another side. What other "side-side ratios" are impossible?

Here are the impossible ratios I've come across in researching this: 1/2, 2/5, 2/3, 62/63, 6/7

There seem to be infinitely more from the generation method given by Fine in his "On Rational Triangles" paper. But, I would like a general check for any given ratio. For example, I've used Matlab to evaluate this ratio for millions of triangles and never seen the ratios 1/3, 1/4, or 1/5; so, does anybody know if these are truly impossible?

I have a hunch that these impossible ratios have a neat pattern...so please help! Thanks!

ADDED 1/27/2012: From the answers of Alan and Jamie, I've learned that most ratios can quickly be checked with Sage (using descents in elliptic curves): https://picasaweb.google.com/107800252627134603876/Math#5702303639089321090 From the plotted summary, you can see that all the ratios I mentioned truly are impossible. I'm guessing that the few uncheckable ratios in the plot could be checked with some new creative descent method if that were one's lifetime goal... it even sounds like Magma already has the capability to clarify some of these points.

Answer 1

Suppose the ratio is called $k$, and let the sides be $g, kg, h$, with the angle between sides $g$ and $kg$ called $A$, with all these variables rational.

Then $\Delta = \frac{1}{2}kg^2 \sin A$, so if the triangle is a Heron triangle with rational area, then $\sin A$ must be rational. Also, by the cosine rule $\cos A$ must be rational.

We can set $\cos A=(1-t^2)/(1+t^2)$ and $\sin A=2t/(1+t^2)$, and $t$ must be rational.

For $h$ to be rational, we must have $k^2-2k\cos A+1$ a rational square, and substituting the rational form for $\cos A$ means that the quartic

$(k+1)^2t^4+2(k^2+1)t^2+(k-1)^2$

must be a rational square. For fixed $k$, the value $t=0$ gives a solution, so this quartic is birationally equivalent to an elliptic curve. Setting $k=m/n$, with $m,n \in \mathbb{Z}$ and doing some standard algebra, leads to the elliptic curve

$W^2=Z(Z-m^2)(Z-n^2)$

with $t=W/(Z(m+n))$.

The curve has (usually) only $3$ finite torsion points at $(0,0)$, $(m^2,0)$ and $(n^2,0)$, none of which give a non-trivial value of $t$. Thus, to get solutions, we need the rank of the elliptic curve to be greater than $0$.

Testing gives the rank equal to $0$ for $k=1/2,1/3,1/4,1/5$ so no solutions will exist. For $k=1/6$, however, the rank is $1$, with generator $(81,540)$. This gives $t=20/21$, and, after clearing denominators a triangle with sides $174, 29, 175$ which has area $2520$.

Answer 2

Bob, you are correct, there is no known algorithm guaranteed to compute the rank of an elliptic curve. There are however very effective algorithms for obtaining an upper bound on the rank by computing a Selmer group.

This can be done in either Sage or Magma. Since Sage is free, I will put some sample code here:

sage: k = 6/17
sage: E = EllipticCurve([0,-1 - k^2, 0, k^2, 0]).minimal_model()
sage: E.selmer_rank() - E.two_torsion_rank() # Upper bound from 2-Selmer Group
0    
sage: E.rank_bound() # Computes a sometimes better bound
0

In these cases the 2-Selmer group is relatively easy to compute, because every point of order 2 on the elliptic curve is rational. It's behavior is determined, in a somewhat complicated way, by the prime factorization of $n*m*(n+m)*(n-m)$ and whether those primes are squares modulo one another.

The going conjecture is that if you pick $k$ "at random", the the rank of the elliptic curve Allan wrote down has about a 50% chance of being 0 and a 50% percent chance of being 1.

Unfortunately, some results of Gang Yu, and certain elliptic curve analogues of Malle's corrections to the Cohen-Lenstra heurstics, suggest that if you pick $k$ at random and compute the upper bound you will quite often get an upper bound of 2, or 4, or 6, or so on, even when the rank is 0. If you want to do something more exhaustive, you might need to do so-called higher descents. These are, at present, best done with Magma.

EDIT STARTS HERE:###

For simplicity, I'm changing coordinates a bit starting from $W^2 = Z(Z-m^2)(Z-n^2)$ let $x = Z/m^2$ and $y = W/m^3$ then we see from Allan's answer above that a Heron triangle with side lengths $a, ka, b$ will give rise to a rational point on the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ other than $(0,0)$, $(1,0)$, or $(k.0)$.

The Mordell-Weil theorem (since we are working over $\Bbb Q$ I should probably call it Mordell's theorem) tells us that the group of rational points $E_k(\Bbb Q)$ is a finitely generated abelian group, thus isomorphic to $T_k \times \Bbb Z^{r_k}$ for some finite abelian group $T_k$ and some non-negative integer $r_k$.

Given an elliptic curve, the torsion subgroup is easy to compute in Sage:

sage: k = 1/3
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: E.torsion_subgroup().invariants()
(2, 2)

but let me tell you what $T_k$ is for all $k$. Since we know that are $T_k$ contains the 3 points $(0,0)$, $(1,0)$, and $(k,0)$ of order 2, a deep theorem of Mazur assures us that $T_k$ is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z /2j\Bbb Z$ where $j = 1, 2, 3,$ or $4$.

I'm going to use the fact that Noam mentioned in the comments that $E_k$ precisely the quadratic twists by $-1$ of those elliptic curves with torsion subgroup $\Bbb Z/2 \Bbb Z \times \Bbb Z/4 \Bbb Z$. In particular, this means that there is a rational cyclic 8-isogeny among those elliptic curves isogenous to $E_k$ over $\Bbb Q$. If we had $j = 3$ then each of those isogenous curves would also possess a rational cyclic 3-isogeny. So there would be a rational cyclic 24-isogeny among these curves. This does not happen over $\Bbb Q$.

No let's consider the possibility of $j = 2$ or $j = 4$. If $E_k$ possesses a point of order 4 over $\Bbb Q$ then $E_k$ would have to possess 3 points of order 4 over $\Bbb Q(i)$. This will happen if and only if $k^2 - 1$ is a square up to sign. See for example:

sage: k = 3/5
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()    
sage: k^2 - 1
-16/25
sage: E.torsion_subgroup().invariants()
(2, 4)
sage: k = 5/4
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: k^2 - 1
9/16
sage: E.torsion_subgroup().invariants()
(2, 4)

The condition that $E_k(\Bbb Q(i))$ contains 3 points of order 4 rules out the possibility of a point of order 8. Since I've already used Mazur's theorem classifying torsion subgroups over elliptic curves over $\Bbb Q$. I'll go ahead say that Kamienny's theorem classifying torsion subgroups of elliptic curves over quadratic fields and say that $E_k(\Bbb Q(i))$ cannot have 3 points of order 4 and a point of order 8. You rule out $j = 4$ in a simpler way that comes down to Fermat's $a^4 + b^4 = c^2$, but I won't do that now.

So here is the best answer I can give to your question at present:

Let $k$ be a rational number. If there are infinitely many non-congruent Heron triangles with sides of the form $a, ka, b$ if then the Mordell-Weil rank of the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ is positive. If the Mordell-Weil rank of $E_k$ is zero, then there can, and will be finitely many congruence classes of Heron triangles with side lengths of the form $a, ak, b$ if and only if $k^2 - 1$ is a square up to sign. In particular one can take the right triangle with sides $1, k, \sqrt{|k^2 -1|}$.

Answer 3

I did some calculations this morning on the elliptic curve

$W^2=Z(Z-m^2)(Z-n^2)$

I have several codes in Ubasic or Pari which use the Birch & Swinnerton-Dyer conjecture to predict ranks.

For $1 \le m \le n \le 99$ with $\gcd(m,n)=1$, the results suggest 645 curves have rank 0, 730 curves have rank 1 and 126 curves have rank greater than 1.

The codes also use the BSD conjecture to predict the height of the generator of the rank 1 curves. For this range, the highest height comes from m=2 n=81, with height 34.9/69.8 (depending on which height normalisation you like). m=1 n=83 gives a higher BSD value but I found a point with height a quarter of this value.

For (2,81), the triangle produced has integer sides with over 30 digits.

Allan MacLeod

Answer 4

Starting with $Z(Z - m^2) = (Z - n^2) b^2$, to which Allan MacLeod's elliptic curve can be reduced by taking $W = (Z - n^2)b$, one can find a general parametrization of m, n as follows.

Letting $a, x = Z/b, m/b$ gives $a^2 - a b x^2 = a b - n^2$, in which then letting $n = a y$ gives $b (x^2 + 1) = a (y^2 + 1)$. This implies $x^2 + 1, y^2 + 1 = a z, b z$ for some rational $z$, and multiplying these gives after composition $(\frac{x y + 1}{z})^2 + (\frac{x - y}{z})^2 = a b$.

Letting $a, b = k A, k B$, where $A, B$ are coprime integers, the factor $k^2$ in the preceding equation can be absorbed into each square, and we can conclude that $A, B$ are each a sum of two squares, say $p^2 + q^2, r^2 + s^2$ resp, so that $B (x^2 + 1) = A (y^2 + 1)$ becomes $(p x + q)^2 + (p - q x)^2 = (r y + s)^2 + (r - s y)^2$

The latter has general solution as follows, for rational $u, v$ with $u^2 + v^2 = 1$ :

$p x + q, p - q x = u (r y + s) + v (r - s y), v (r y + s) - u (r - s y) w$

So that:

$ x = \frac{u (r y + s) + v (r - s y) - q}{p} = \frac{p - v (r y + s) - u (r - s y)}{q}$

which expresses $y$ and then $x$ rationally in terms of $u, v$ and $p, q, r, s$ (and the latter appear homogenously, so one of them is disposable i.e. can be assumed equal to 1).

edit: I should clarify that this isn't a rational parametrization of one elliptic curve, which of course is impossible if $m^2, n^2$ are non-zero and distinct. What it does is start with a supposed "symbolic" solution and express the roots parametrically in a way consistent with that solution. In other words it constructs a multi-dimensional pencil of all elliptic curves having the required form.

Answer 5

With reference to Noam's result, equivalent to $a(a - x^2) = (a - 1) y^2$, it isn't hard to show that this is rational, and $a, x, y$ can be expressed in terms of two parameters.

From this one can churn out possible values of $x$ by the bushel. Deciding whether any given value of $x$ is possible is harder, and may be no more practical using this result than those already discussed.

Anyway, I'll sketch how the solution is obtained.

Firstly, the N&S condition for the above, considered as a quadratic in $a$, to have rational $a$ when $x, y$ are rational is $(x^2 + y^2)^2 - 4 y^2 = z^2$ for some rational $z$.

This implies some rational $p$ for which:

$$\frac{y}{x^2 + y^2} = \frac{p}{p^2 + 1},\qquad \frac{z}{x^2 + y^2} = \frac{p^2 - 1}{p^2 + 1}$$

Reciprocating the first and completing squares expresses it in the form:

$$x^2 + (y - \frac{p^2 + 1}{2 p})^2 = (\frac{p^2 + 1}{2 p})^2$$

so there must be some rational $q$ for which:

$$y = \frac{p^2 + 1}{2 p} + \frac{q^2 - 1}{2 q} x, \qquad \frac{p^2 + 1}{2 p} = \frac{q^2 + 1}{2 q} x$$

From the last one can express $x = x(p, q)$, and plugging this into the preceding equation gives $y = y(p, q)$. Finally, we can obtain $z$ and then $a$ also both in terms of $p, q$.

NOTE: As the PC I am typing this on doesn't format equations, I am "flying blind". So if the result looks hideous due to any missing dollars and suchlike, perhaps someone could edit it.