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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.

Bumblebee
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mr_T
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  • 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 09:12
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    $90^\circ=\frac\pi2$ – Quang Hoang Sep 30 '15 at 09:13
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    @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 09:18
  • @QuangHoang oops.. thanks! – mr_T Sep 30 '15 at 09:18
  • I've edited the question considering the comments – mr_T Sep 30 '15 at 09:20
  • The triangle with sides $2ab$, $a^2-b^2$ and hypothenus $a^2+b^2$ is a Pythagorean triple for $a>b$. Different choices of a, b will let you experiment with angles. – Paul Sep 30 '15 at 09:29
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    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
  • @mr_T I think the playful way you phrased the question has confused the people trying to answer. You could have just said, "Do there exist any Pythagorean triples such that the corresponding triangle has angles that are rational multiples of pi?" – Jerry Guern Sep 30 '15 at 13:42

4 Answers4

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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 $\Bbb Q(i)$.

Lubin
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  • 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 01 '15 at 14:29
  • I wish all my errors were as minor! – Lubin Oct 01 '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
  • Well, @JosiahYoder, any root of unity has to be an integral element of the ring, i.e. root of a monic polynomial with coefficients from $\Bbb Z$. The set of all these in $\Bbb Q(i)$ is the ring of Gaussian integers, numbers of form $m+ni$. Multiply this by its conjugate and you get $m^2+n^2$, which has to be a unit of $\Bbb Z$, thus $\pm1$, but of course $-1$ is out of the question. Now how many integer solutions of $m^2+n^2=1$ are there? Just the obvious four. And there you are. – Lubin Apr 26 '18 at 19:54
  • @Lubin Thank you. I guess I'm still struggling with how we know the only algebraic integers in Q(i) are Gaussian integers – Josiah Yoder Apr 26 '18 at 20:40
  • Indeed, @JosiahYoder, for a general algebraic number field, it’s definitely a nontrivial problem. Even for quadratic fields $\Bbb Q(\sqrt d,)$, the answer is different, depending on the square-free integer $d$ you’re dealing with. If $d\equiv1\pmod4$, the integers are all of form $(m+n\sqrt d,)/2$ with $m$ and $n$ of the same parity; otherwise it’s just the numbers of form $m+n\sqrt d$. Write down the minimal polynomial of $\lambda+\mu\sqrt d$ with $\lambda,\mu$ rationals, and you’ll rapidly see that $2\lambda\in\Bbb Z$, and the rest is not hard to prove. – Lubin Apr 26 '18 at 21:01
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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.

Piquito
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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.

Lucian
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  • 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
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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.

N. S.
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