# Is there a pythagorean triple such that all angles of the corresponding triangle are simple fractions of $\pi$?

Obviously, the most interesting pythagorean triple $(a, b, c)$ would be one for which the corresponding triangle (with integer side lengths $a, b, c$) has angles 90°, 60° and 30° ($\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{\pi}{6}$). This would mean that $c = 2a$ (since $\sin \frac{\pi}{6} = \frac{1}{2}$).

But in case this doens't exist, I would be interested to learn about any triple leading to "nice" or "interesting" acute angles.

• What do you want? Do you want a triplet other than $(\frac\pi2,\frac\pi3,\frac\pi{6})$? Or you are thinking that $(\frac\pi2,\frac\pi3,\frac\pi{6})$ doesn't make up a right triangle? Clarify. And also, a pythagorean triplet is composed of side lengths, not angles. – Aditya Agarwal Sep 30 '15 at 9:12
• $90^\circ=\frac\pi2$ – Quang Hoang Sep 30 '15 at 9:13
• @AdityaAgarwal I want to have a pythagorean triple (thus side lengths a, b, c all integer) that also has certain angle properties. – mr_T Sep 30 '15 at 9:18
• @QuangHoang oops.. thanks! – mr_T Sep 30 '15 at 9:18
• This is really a fascinating question. If no example of this exists, I'd really like to see a proof. – Jerry Guern Sep 30 '15 at 10:30

This is a consequence of the fact that the unit group of $\Bbb Z[i]$ is $\{\pm1,\pm i\}$.

For, take a Pythagorean triple: $a^2+b^2=c^2$, with the angle $\beta$ opposite the side of length $b$. You are asking whether it is possible for $n\beta\equiv0\pmod{2\pi}$.

Now $z=\frac a c+\frac b ci$ is an element of $\Bbb Q(i)$ on the unit circle, and its argument is $\arctan(b/a)=\beta$. The argument of $z^n$ is $n\beta\pmod{2\pi}$.

But if $n\beta\equiv0\pmod{2\pi}$, then $z^n=1$, and so $z$ is a root of unity, of which there are only the powers of $i$ in $\Bbb Q(i)$.

• If I am not wrong, you really and involuntarily prove Niven's theorem (of which I do not know the original proof). Very good. – Piquito Sep 30 '15 at 14:06
• I don’t believe so. Niven’s Theorem is about sines. What I observed (surely it’s not my proof) is about sine and cosine being simultaneously rational. Much, much weaker. – Lubin Sep 30 '15 at 16:51
• Errare humanum est. Regards. – Piquito Oct 1 '15 at 14:29
• I wish all my errors were as minor! – Lubin Oct 1 '15 at 14:31
• How do we know that {±1,±i} are the only roots of unity in Q(i)? – Josiah Yoder Apr 26 '18 at 17:41

Your three angles are rational multiples of $\pi$. If your three sides are also rational $($since this is what the expression Pythagorean triple implies$)$, then this would mean that the sine and cosine of rational multiples of $\pi$ are also rational. But, according to Niven's theorem, this can only happen for the $30^\circ-60^\circ-90^\circ$ triangle.

• When I say to myself, "Oh, God! The triangle with angles 90, 60 and 30 degrees is not Pythagorean”, as suggested Mr_T's post, and I was going to write about it, I read the interesting responses of @Lucian and N. S. Consequently the answer to the post is NONE PYTHAGOREAN TRIANGLE AS DESIRED. – Piquito Sep 30 '15 at 12:41

Let $a,b,c$ the sides of a pythagorean triangle with correlative angles $\alpha$ and $\beta$ ($\gamma$ is always $\frac{\pi}{2}$). It is known, by trascendental number theory, that $\sin\space x$ and $\cos\space x$ are transcendental when $x\ne 0$ is algebraic. Consequently $\alpha$ and $\beta$ are necessarily transcendental because $\sin\space \alpha=\frac{a}{c}$ and $\sin\space \beta=\frac{b}{c}$.

Besides $\alpha + \beta= \frac{\pi}{2}$ but this do not mean necessarily that $\alpha$ and $\beta$ are rational fractions of $\pi$.

Your question is interesting. I think the most probably, with exception of $(\alpha, \beta)$=$(\frac {\pi}{3},\frac{\pi}{6})$ is that$(\alpha, \beta)$= $(\frac{\pi}{2}-h,\frac {\pi}{2}+h)$ for some "disturbed" $h$.

I'll try to get a better answer.

NOTE.-The triangle with angles 90, 60 and 30 degrees is not Pythagorean. There are consequently no exception for pythagorean triangles. And there is just one exception for rectangle triangles according to the Niven's Theorem cited in other answer here by @Lucian.

Let $A$ be the angle which is $\frac{\pi}{2}$ and assume that another angle $B=\frac{m}{n}\pi$.

Then $2 \cos(B) =e^{\frac{m}{n}\pi i}+e^{-\frac{m}{n}\pi i}$ is an algebraic integer, as the sum of two algebraic integer. This implies that $2 \cos(B)$ is an integer, and since we are in the first quadrant we have $$2 \cos(B) \in \{ 0 ,1, 2\}$$

It is easy to check all three cases.