I'm wondering if a triangle with integer sides and integer angles in degrees exists, which is not equilateral. Could someone come up with a proof for why or why not?
Edit: Please do not answer with a triangle which has integer sides only. The triangle must have integer angles in degrees; this means they must actually be integers, not rounded off.
Best Answer
As mentioned on the Wikipedia page for "integer triangle", the law of cosines forces the cosine of an angle in a triangle with integer sides to be a rational number.
Now, when does an integer angle (in degrees) have a rational cosine? We can go one step further: when does a rational angle (in degrees) have a rational cosine? This question is answered with a clever proof on page 2 of When is the (co)sine of a rational angle equal to a rational number? by Jörg Jahnel. The answer is that the rational cosines of the rational angles are just $\pm1,\pm\frac12,0$. (This is known as Niven's Theorem and an alternate proof can be found at ProofWiki.)
Since cosines of $\pm1$ wouldn't correspond to the angle of a genuine triangle, the only potential options are $\cos\theta=0,\pm\frac12$ for angles of $\theta=60^\circ,90^\circ,120^\circ$. But if there is one angle of $90^\circ$ or $120^\circ$, there isn't enough room in $180^\circ$ for two more angles of at least $60^\circ$. And the only triangles whose angles are all $60^\circ$ are equilateral.
The answer to the question "Can a non-equilateral triangle with integer sides and integer angles (in degrees) exist?" is no.