Heronian triangles and Somos-5 sequences? (Part 2)

Question

(This continues this post, but asks if we can find another family of solutions.)

Given the Heronian triangle with half-sides $(a,b,c)$, area $A$, and medians $(m_1,m_2,m_3)$,

$$A=\sqrt{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}$$ $$m_1=\sqrt{2 (b^2 + c^2) - a^2}$$ $$m_2=\sqrt{2 (c^2 + a^2) - b^2}$$ $$m_3=\sqrt{2 (a^2 + b^2) - c^2}$$

Using an elliptic curve, one can find non-zero integer $(a,b,c,m_1,m_2,m_3)$ but non-rational $A\neq0$. If we start with area $A$, Schubert conjectured in 1905 that solving the first three equations in non-zero integers was impossible, but was proven wrong in 1989.

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I. Three quadratics

We do the cyclic substitution $(a,b,c) = (p+q,\,q-r,\,r-p)$, then the first three equations simplify to,

\begin{align} 16\, p\, q\, (-p - q + r) r\, &=\, A^2\\[5pt] (p - q)^2 - 4 (p + q) r + 4 r^2 &= m_1^2\\[5pt] (2 p + q)^2 - 2 (2 p - q) r + r^2 &= m_2^2 \end{align}

So three quadratics in $r$ to be made squares. They are infinitely solvable.

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II. Examples

In Part 1, using two Somos-5 sequences, one can find infinitely many primitive solutions to the three quadratics. For example, starting with $(p,q,r) = (24,\,49,-2)$ from the family, then,

$$(a,b,c) = (p+q,\,q-r,\,r-p) = (73, \,51,\, -26)$$

Ignoring signs, this yields the smallest integer area $A=1680$ and,

$$(m_1, m_2, m_3) = (35,\, 97,\, \sqrt{u})$$

The transformation also goes in the other direction. Given $(a,b,c)$ not generated by the Somos-5 sequences, for example the primitive,

$$(a,b,c) = (4368, 1241, 3673)$$ $$(m_1, m_2, m_3) = (3314, 7975,\, \sqrt{v})$$

We can use this $(a,b,c)$ to find,

$$(p,q,r) = (-273, 4641, 3400)$$

which also solves the three quadratics, suggesting there may be another family.

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III. Question

Q: Can we find another infinite family of $(p,q,r)$ that solves the three quadratics above, distinct from the one using Somos-5 sequences? Perhaps using an elliptic curve or other recurrence relations?

Answer 1

(An analysis/comment of known data that may lead to an answer.)

I. The first family

In Andrew Hone's 2022 paper, formulas for $(a,b,c)$ were found using two Somos-5 sequences,

$S_n = 1,\, 1,\, 1,\, 2,\, 3,\; 5,\, 11,\, \color{blue}{37},\; \color{blue}{83},\dots$

$T_n = 0, 1, -1, 1, 1, -7, 8, -1, \color{blue}{-57},\dots$

In Part I of this question, Somos found that the $(a,b,c)$ shared a generalized Somos-9 recurrence relation,

$$a_0\, a_9 =-\color{blue}{37}a_1\,a_8+(\color{blue}{57}/3)(\color{blue}{83})\big(105 a_2\, a_7 - 111 a_3\, a_6 + 1628 a_4\, a_5\big)$$

For aesthetics, we choose the initial conditions,

$a_n = -875,\, 51,\; 2, -1, 1,\; 0, -1,\, 73,\, -626\\[5pt] b_n = -626,\, 73, -1,\, 0,\, 1, -1,\, 2,\, 51,\, -875\\[5pt] c_n \,= \;291,\; 26, -1,\, 1,\, 0, -1,\, 1, -26, -291$

Notice that $a_n$ and $b_n$ are palindromic with each other, while $c_n$ is palindromic (unsigned) with itself, a phenomenon explained by Somos in his answer. Plugging these values into the recurrence, one finds the next solution,

$$(a_9,b_9,c_9) = (-28779, 13816, 15155)$$

and so on for all $n$.

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II. A second family?

Assuming the recurrence,

$$a_0\, a_9 =P\,a_1\,a_8 + Q\, a_2\, a_7 + R\, a_3\, a_6 + S\, a_4\, a_5$$

and knowing at least four non-trivial solutions $(a,b,c)$ with area $A\neq0$ and five trivial ones where $A = a\pm b \pm c =0$, one can solve for $(P,Q,R,S)$ as a system of four equations in four unknowns.

The tricky part is choosing the correct sign since $(\pm a, \pm b)$ are both valid. However, using the palindromic property and since there are only 8 non-zero $a_n$, plus shifting indices 4 times, then one only has to consider $2^{12}=4096$ possibilities, easily done by Mathematica. With that, I managed to recover Somos' $(P,Q,R,S)$.

Randall Rathbun has a database of the 8 smallest primitive solutions $(a,b,c)$. Four of these belong to the first family, but it is unknown if there is a second family for the other four namely,

$a_n = 1241, 14384, 2288232, 22816608\\[5pt] b_n = 4368, 14791, 1976471, 20565641\\[5pt] c_n = 3673, 11257, 2025361, 19227017$

Using the same method to recover Somos' integer $(P,Q,R,S)$, Mathematica did not find any. Even changing the order of the five trivial $(a,b,c)$ still did not yield a result. So either I made a coding error, or its recurrence is really not of the same form.

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III. Factorizations

The pattern for the new $(a,b,c)$ seems hard to crack. For example $c_1 = 3673$ is prime. But if we do the same substitution in Part I, namely $(a,b,c) = (p+q,\,q-r,\,r-p)$, then solving for $(p,q,r)$,

$$(p,q,r) = \frac{a - b - c}2,\, \frac{a + b + c}2,\, \frac{a - b + c}2$$

we get,

$(p_1,q_1,r_1) = (-2^3\times 5^2\times 17,\; 3\times 7\times 13\times 17,\; 3\times 7\times 13)$

$(p_2,q_2,r_2) = (-2^3\times 3^6,\; 2^3\times 7\times 19^2,\; 5^2\times 7\times 31)$

$(p_3,q_3,r_3) = (-2^5\times 3^2\times 5^2\times 7\times 17,\; 2^3\times 3^2\times 11^2\times 19^2,\; 23^2\times 47^2)$

$(p_4,q_4,r_4) = (-5^2\times 7^2\times 13^2\times 41,\; 17\times 23^2\times 59^2,\; 2^4\times 3\times 11^2\times 43^2)$

Similar factorization behavior can be found in the first family, suggesting both are products of terms of a Somos-type sequence. In the first case, the primes $37$ and $83$ stood out, identifying the Somos-5. In the second case, it is harder to identify which sequence it comes from, if any.