Herons triangles with integer area, sides, and at least two medians

Question

Known eight primitive Heron triangles with integer area, sides, and two medians, see A181928. Let $(a,b,c)$ be half-sides, $s$ is area, and $(m_a,m_b,m_c)$ be medians.

I. General properties

$$s=\sqrt{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}$$ $$m_a=\sqrt{2 (b^2 + c^2) - a^2}$$ $$m_b=\sqrt{2 (c^2 + a^2) - b^2}$$ $$m_c=\sqrt{2 (a^2 + b^2) - c^2}$$

We are looking for integer $(a,b,c)$ such that $s$ and two of the $m_k$ are also integers.

II. Other properties

$$9a^2=2 (m_b^2 + m_c^2) - m_a^2$$ $$9b^2=2 (m_c^2 + m_a^2) - m_b^2$$ $$9c^2=2 (m_a^2 + m_b^2) - m_c^2$$ $$m_a^2 - 3 a^2 = 2 (3 b^2 - m_c^2) = 2 (3 c^2 - m_b^2) = x$$ $$m_b^2 - 3 b^2 = 2 (3 c^2 - m_a^2) = 2 (3 a^2 - m_c^2) = y$$ $$m_c^2 - 3 c^2 = 2 (3 a^2 - m_b^2) = 2 (3 b^2 - m_a^2) = z$$ $$x+y+z=0$$ $$m_a^2 + 3 a^2 = m_b^2 + 3 b^2 = m_c^2 + 3 c^2 = w$$ $$m_a^2 + m_b^2 + m_c^2 = 3 (a^2 + b^2 + c^2)=\frac{3}{2}w$$

where $(x,y,z,w)$ is some integer.

Exact two half-sides is odd and one is even, exact two medians is odd and one is even.

Then integer $w$ is of form $(km \pm 3ln)^2+3(kn \mp lm)^2=(k^2+3l^2)(m^2+3n^2)$, where $(k,l,m,n)$ is any positive integer.

III. Relations

Some relations between known primitives Herons triangles, selected by colors.

Note: By common convention, $p|q$ means "$p$ divides $q$". In the second example below, $97|297$ just means $297=3\times97$ and $(a,b,c) = (626,875, 297)$. For simplicity, the non-integer $m_i$ is denoted as $\sqrt{n_k}$.

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1. $s=1680$

$a=26, b = 73, c = 51$

$m_a= \sqrt{n_1}, m_b = 35, m_c = \color{red}{97}$

$x = 13156, y = -14762, z = 1606, w = 17212$

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2. $s=221760$

$a = 626, b = 875, c = \color{red}{97}|291$

$m_a = 1144, m_b = \color{blue}{433}, m_c = \sqrt{n_2}$

$x = 133108, y = -2109386, z = \color{red}{97}|1976278, w = \color{red}{97}|2484364$

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3. $s = 8168160$

$a = 3673, b = 4368, c = 1241$

$m_a = \sqrt{n_3}, m_b = 3314, m_c = 7975$

$x = -12724706, y = -46255676, z = 58980382, w = 68220868$

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4. $s = 95726400$

$a = 13816, b = 28779, c = \color{blue}{433}|15155$

$m_a = \color{magenta}{21937}|43874, m_b = \color{red}{97}|3589, m_c = \sqrt{n_4}$

$x = 1352282308, y = -2471811602, z = \color{blue}{433}|1119529294, w = \color{blue}{433}|2497573444$

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5. $s = 302793120$

$a = 11257, b = 14791, c = 14384$

$m_a = \sqrt{n_5}, m_b = 21177, m_c = 22002$

$x = 344466078, y = -207855714, z = -136610364, w = 1104786372$

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6. $s = 569336866560$

$a = 1823675, b = \color{magenta}{21937}|1930456, c = 185629$

$m_a = \color{blue}{433}|2048523, m_b = \sqrt{n_6}, m_c = \color{orange}{13\cdot96181}|3751059$

$x = -5780925035346, y = \color{magenta}{21937}|(-8186144209212), z = 13967069244558, w = \color{magenta}{21937}|14173817998404$

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7. $s = 8548588738240320$

$a = \color{orange}{13\cdot96181}|46263061, b = 2442655864, c = 2396426547$

$m_a = \sqrt{n_7}, m_b = \color{magenta}{21937}|2350198558, m_c =\color{#0a0}{661\cdot107581}|2488886435$

$x = \color{orange}{13\cdot96181}|23410294646947500526, y = -12376269747775480124, z = -11034024899172020402, w = \color{orange}{13\cdot96181}|23423136271826038852$

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8. $s = 17293367819066194215360$

$a = 31982445133, b = 356388643246, c = \color{#0a0}{661\cdot107581}|336426334971$

$m_a = \color{orange}{13\cdot96181}|692364218455, m_b = 318430912888, m_c = \sqrt{n_8}$

$x = 476299580606746896423958, y = -279640348821488939749004, z = \color{#0a0}{661\cdot107581}|(-196659231785257956674954), w = \color{#0a0}{661\cdot107581}|482436841386859028750092$

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IV. Question

Is it possible to use these relations to search other Heron triangles with two integer medians?

Answer 1

(New answer)

I. Formulas

As the OP made a crucial typographic slip and formula should have been $s\to s^2$, we make some modifications. But the answer to "... can we use these relations to search other Heron triangles with two integer medians" is YES.

Define the Somos-5 sequence (starting with $n=0$),

$S(n) = 1,\, 1,\, 1,\, 2,\, 3,\, 5,\, 11,\, 37,\, 83,\, 274,\, 1217,\; 6161,\; 22833,\dots$ (A006721)

$T(n) = 0, 1, -1, 1, 1, -7, 8, -1, -57, 391, -455, -2729, 22352,\dots$ (A360381)

then there is a formula for $(a,b,c)$ in a 2022 paper (in page 6) by Andrew Hone based on the work of Buchholz, Rathbun, and Elkies, namely,

$$a = S_{n+1}\,S_{n+2}^3\, S_{n+3}\, T_{n+2} + S_{n}^2\, S_{n+1}\, T_{n+3}\, T_{n+4}^2$$ $$b = S_{n}^2\, S_{n+1}\, T_{n+3}\, T_{n+4}^2 - T_{n+1}\, T_{n+2}^3\, T_{n+3}\, S_{n+2}$$ $$\; c = T_{n+1}\, T_{n+2}^3 \,T_{n+3}\, S_{n+2} - S_{n+1}\, S_{n+2}^3 \,S_{n+3}\, T_{n+2}$$

Based on the relations given by the OP, the variables $(x,y,z,w)$ have the formulas,

$$x = 2(-2a^2 + b^2 + c^2)$$ $$y = 2(a^2 - 2b^2 + c^2)$$ $$z = 2(a^2 + b^2 - 2c^2)$$ $$w = 2(a^2 + b^2 + c^2)$$

II. Examples

For $n=1,2,\dots6$, the formula yields ALL of the "colored" solutions given by the OP. (Out of the 8, only 6 were colored.) To find a bigger example, we compare it to $n=6$,

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$n=6$ with $s = 17293367819066194215360$

$(a,b,c) = (31982445133, 356388643246, \color{#0a0}{661\cdot107581}|336426334971)$

$(m_a,m_b,m_c) = \color{orange}{13\cdot96181}|692364218455, \color{blue}{3618533101}|318430912888, \sqrt{z_6}$

$(x,y,z,w) = (476299580606746896423958, -279640348821488939749004, \color{#0a0}{661\cdot107581}|(-196659231785257956674954)),\color{#0a0}{661\cdot107581}|482436841386859028750092$

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$n=7$ with $s = 74821433268961491419039525520$

$(a,b,c) = (427416063477339, 809817782676175, \color{blue}{3618533101}|387667925242534)$

$(m_a,m_b,m_c) = (\color{#0a0}{661\cdot107581}|1195616172307129, 100682580529477, \sqrt{z_7})$

$(x,y,z,w) = (881444557526956338413004659878, -1957277541393595184733071998346, \color{blue}{3618533101}|1075832983866638846320067338468, \color{blue}{3618533101}|1977551505437744426936403785404)$

Then a divisor of $m_b = 100682580529477$ can integrally divide three terms for $n=8$, and so on for all $n$.

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(Old answer)

I. Formulas

My interpretation of this post is the OP wishes to find integer $(a,b,c)$ and $(x,y,z,w)$ fulfilling certain conditions. The role of the $(a,b,c)$ is to define the area $s$ of Heron's triangle,

$$s=(a + b + c) (b + c - a) (a + c - b) (a + b - c)$$

He then gives the three medians,

$$m_a=\sqrt{2 (b^2 + c^2) - a^2}$$ $$m_b=\sqrt{2 (c^2 + a^2) - b^2}$$ $$m_c=\sqrt{2 (a^2 + b^2) - c^2}$$

The next triplet he gives is identically satisfied by the $(m_a, m_b, m_c)$ above for any $(a,b,c)$,

$$9a^2=2 (m_b^2 + m_c^2) - m_a^2$$ $$9b^2=2 (m_c^2 + m_a^2) - m_b^2$$ $$9c^2=2 (m_a^2 + m_b^2) - m_c^2$$

To find $(x,y,z)$, one simply substitutes these $(m_a, m_b, m_c)$ into the next relations,

$$x = m_a^2 - 3 a^2 = 2 (3 b^2 - m_c^2) = 2 (3 c^2 - m_b^2) $$ $$y = m_b^2 - 3 b^2 = 2 (3 c^2 - m_a^2) = 2 (3 a^2 - m_c^2) $$ $$z = m_c^2 - 3 c^2 = 2 (3 a^2 - m_b^2) = 2 (3 b^2 - m_a^2) $$

To get,

$$x = 2(-2a^2 + b^2 + c^2)$$ $$y = 2(a^2 - 2b^2 + c^2)$$ $$z = 2(a^2 + b^2 - 2c^2)$$

and which obeys $x+y+z=0$ as expected. To find $w$, we again substitute $(m_a, m_b, m_c)$ into the relations,

$$w = m_a^2 + 3 a^2 = m_b^2 + 3 b^2 = m_c^2 + 3 c^2 \quad$$

to get,

$$w = 2(a^2 + b^2 + c^2)$$

which obeys the last equation specified by the OP,

$$m_a^2 + m_b^2 + m_c^2 = 3 (a^2 + b^2 + c^2)=\frac{3}{2}w$$

so all variables $(a,b,c)$ and $(x,y,z,w)$ are accounted for.

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II. Examples

We use the OP's first example,

$$(a,b,c) = (26,\,73,\,51)$$

Using the formulas above, we recover his,

$$(x,y,z,w) =(13156,\, -14762,\, 1606,\, 17212)$$

and his medians,

$$(m_a, m_b, m_c) = (4\sqrt{949},\, 35,\, 97)$$

But if the OP wishes that the medians are all integers as well, a simple solution exist. (Edit: We revise the answer to avoid $a+b=c$ and Empy2's comment about $s=0$.) To solve the simultaneous,

$$m_a^2 = 2 (b^2 + c^2) - a^2 $$ $$m_b^2 = 2 (c^2 + a^2) - b^2 $$ $$m_c^2 = 2 (a^2 + b^2) - c^2 $$

Do the substitution,

\begin{align} a &= p + q + (p - q) r\\[5pt] b &= 2 (p r - q)\\[5pt] c &= -p + q + (p + q) r\end{align}

to get,

\begin{align} m_a^2 &= (p - 3 q)^2 - (6 p^2 + 16 p q - 6 q^2) r + (3 p + q)^2 r^2\\[5pt] m_b^2 &= 4(p+qr)^2\\[5pt] m_c^2 &= (p + 3 q)^2 + (6 p^2 - 16 p q - 6 q^2) r - (3 p - q)^2 r^2 \end{align}

The second is now a square. Two quadratic polynomials to be made squares and with a rational point can be expressed as an elliptic curve. One solution is $(p,q,r) = (5,2,-46/5)$ which, after scaling, yields,

$$\quad (a,b,c) = (103, 480, 337)$$ $$(m_a, m_b, m_c) = (823, 134, 607)$$

From this initial $(p,q,r)$, infinitely more can be found, hence infinitely more integer $(a,b,c)$ with integer medians $(m_a, m_b, m_c)$.

(Note: See "new answer" at the top where $s\to s^2$ as pointed out by the OP.)