0

(Let an integer triangle be a triangle with only integer side lengths) I have been trying to find all pairs $m$ and $n$ so that both

  1. The triangle with side lengths $m$ and $n$ with a $60$ degree angle in between,
  2. The triangle with side lengths $m$ and $n$ with a $120$ degree angle in between, are integer triangles.

Currently, I believe that there are none, although I cannot show it.

From Law of Cosines, we seek integers $a, b, m, n$ so that \begin{align*} m^2 + mn + n^2 = a^2 \\ m^2 - mn + n^2 = b^2 \end{align*} We can try solving for $m$ and $n$ by solving for $m^2 + n^2$ and $mn,$ then $m + n,$ and finally $m$ and $n$. We get $$m, n = \sqrt{\frac{3a^2 - b^2}{8}} \pm \sqrt{\frac{3b^2 - a^2}{8}}.$$ This means that we need $3a^2 - b^2$ and $3b^2 - a^2$ to be twice perfect squares. Then, I let them be $2x^2$ and $2y^2,$ and then continued to no avail.

I also made a python program that checked when $1 \leq m, n \leq 2000,$ which yielded no solutions, so there are probably no solutions in general.

I would prefer a solution that only uses basic number theory techniques (mods, basic algebra, etc.) although more advanced solutions are welcome too.

invisible
  • 17

1 Answers1

3

Let $x=a^2+b^2$ and $d=ab$. The assumption is that both $x+d$ and $x-d$ are squares. But $x+2d = (a+b)^2$ and $x-2d=(a-b)^2$ are clearly squares. And it is known (and proved on this web page—see Case 2) that $x-2d,x-d,x+d,x+2d$ can never simultaneously be perfect squares when $d\ne0$. So the only solutions to the posted problem are when $a=0$ or $b=0$.

Greg Martin
  • 92,241