# Triangles of the form $a^n+b^n=c^n$

Let $\Delta$ be a triangle with (real) sides length $a,b,c>0$. From this question, we know that $$a+b=c \qquad \iff \qquad \Delta \text{ is a straight line}$$ Now, we know from the Pythagorean theorem that $$a^2+b^2=c^2 \qquad \iff \qquad \Delta \text{ is a right triangle}$$ I'm wondering if it is possible, for a fixed $n>2$, to have a geometric characterization of the triangles which satisfy $a^n+b^n=c^n$, that is: $$a^n+b^n=c^n \qquad \iff \qquad \Delta \text{ is ?}.$$

Notes: For $n>1$, and any $a,b>0$ there exists always a non-degenerated triangle $\Delta$ with $c= (a^n+b^n)^{1/n}$. It seems that for the case $n=3$, things are already more complicated. All the following triangles satisfy $$a^3+b^3=c^3.$$

• In general at least one of $a, b,c$ must be irrational otherwise this would contradict Fermat's Last Theorem – kingW3 Apr 17 '17 at 21:41
• Well, note that you've got a correspondence between these triangles and right-angled triangles via the map $x\to x^{n/2}$. As this map is convex for $n>2$, I'm convinced your triangles will never have angles larger than $\pi/2$. – yo' Apr 17 '17 at 22:16
• I don't think number-theory (or even elementary-number-theory) is an appropriate tag for this. Removed. – Aryabhata Apr 17 '17 at 22:57

Well, for any $m > 0$ there will exist $c^n = m$ and for any $j + k = m; j>0; k>0$ there will exist $a^n = j; b^n = k$. As $(a+b)^n > a^n + b^n = c^n$ we have $a+b > c$ so such triangles exist and such numbers are very common.

• Sorry, I thought it was clear from my answer. I have edited to make it clear. And +1 to you. – Aryabhata Apr 17 '17 at 23:18
• This is a nice observation, but it does not characterizes the triangles for which $a^n+b^n = c^n$. Or am I missing something? – idm Apr 18 '17 at 9:45
• No, it doesn't. But i thought once it was realized that such numbers were common and easy to solve the characteristics of such triangles would seem to be mundane and not worth characterizing. – fleablood Apr 18 '17 at 16:00

If $x = \frac{a}{c}$, $y = \frac{b}{c}$ (note: both are in $(0, 1)$)

then for $n \gt 2$,

$x + y \gt x^2 + y^2 > x^n + y^n = 1$

Thus $a + b \gt c$ and $a^2 + b^2 \gt c^2$ and so $a,b,c$ form the sides of a triangle and the triangle must be acute.

• The implication is one way. Note all acute triangles satisfy the criterion obviously. – Aryabhata Apr 17 '17 at 22:59
• This is a nice observation, but it does not characterizes the triangles for which $a^n+b^n = c^n$. Or am I missing something? – idm Apr 18 '17 at 9:46
• @idm: I would say $a^n + b^n = c^n$ is actually a good characterization already! And no, you are not missing anything. I would not hope for some "elegant" characterization though... We could prove some properties like the triangle is acute etc, but trying to get a simpler characterization might be hopeless (IMO). – Aryabhata Apr 18 '17 at 22:00

If we fix two points of the triangle at $(\pm 1,0)$ (and therefore $c=2$), then we can plot the points satisfying $a+b=2$, $a^2+b^2=2^2$ like this WA link and this WA link.

Likewise, we can plot the cases for $n=3,4,5$ and $10$. ($n=3$, $n=4$, $n=5$, $n=10$)

• Thanks, this is a very nice observation. This suggests somehow that there should exists a Thales theorem holding for ellipses instead of circles. – idm Apr 18 '17 at 20:36