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

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.

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.

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 $10$. ($n=3$, $n=4$, $n=5$, $n=10$)