An integer sided triangle has an area $A$. Heronian triangle areas have no radical, or radical 1. Otherwise, $4 A$ will always be of the form $a\sqrt{r}$, where $r$ is the squarefree radical of the triangle. For the first 24 squarefree numbers, here are the smallest triangles with a given radical.

smallest triangles of a given radical

Now consider sequence $G$.
$$G = (24,8,1,0,8,8,3,0,8,24,1,0)$$

Integer Triangle Radicals conjecture:
For a triangle with radical $r$, let $j = r \mod{12}$.
Then value $4 A/\sqrt{r}$ is divisible by $G_j$ (with $j \in \{1,...,12\})$.

Stated another way, $G_j$ is the GCD of all $4 A/\sqrt{r}$ values corresponding to a particular radical $r$.

Another picture: Triangle Radicals

I've got data from a few million integer triangles suggesting this conjecture is true (supporting code at Triangle Radicals ). All Heronian triangles have area divisible by 6, based on the Brahmagupta parametric form. If this conjecture is true, that hints there might be a more powerful parametric form that classifies all integer triangles.

Does a proof exist?

Mini-challenge:
For radical 953, my scalenes are {{528, 681, 697}, {533, 680, 693}, {561, 656, 689}}.
For radical 977, my scalenes are {{567, 690, 697}, {612, 657, 685}}.
Can someone find triangles with radicals 953 or 977 where $4A$ is divisible by 8, but not 16?

This came out of studying Wheel Graphs with Integer Edges. I noticed all integer-edged wheel graphs with largest edge $<10$ were composed of triangles with exactly 1 or 2 types of radical. Any counterexamples to that?

An integer wheel


Solution 1:

This isn't a full solution, but I think it should help. First, define

$$x = b + c - a, \quad y = a + c - b, \quad z = a + b - c.$$

Then, the area is $\frac{1}{4}\sqrt{xyz(x + y + z)}$. Suppose we are looking at what areas we can get with squarefree part $r$. Now, consider values for $x$, $y$, and $z$ of the form

$$x = r, \quad y = s^2, \quad z = \left(\frac{r + s^2 - 1}{2}\right)^2,$$

so that $x + y + z = \left( \frac{r + s^2 + 1}{2} \right)^2$. Here, $s$ can be any integer whose parity is the opposite of $r$. Then, we are basically looking for possible integer values of

$$s \left( \frac{r + s^2 - 1}{2} \right)\left( \frac{r + s^2 + 1}{2} \right).$$

For large enough primes $p$, it shouldn't be too hard to find values of $s$ such that the above expression is not divisible by $p$. So already we know the desired gcd doesn't contain large prime factors. Then, it remains to check some small cases like $p = 2, 3, 5$.