Don't Any How Define A Cross Product

If we define a cross product as, $a\otimes b=b^a$

consider, 

$\frac{1}{m}\otimes a = a^{\frac{1}{m}}$

 $\frac{1}{n}\otimes b = b^{\frac{1}{n}}$

 $\frac{1}{k}\otimes c = c^{\frac{1}{k}}$

and

$c^\frac{1}{k}=a^{\frac{1}{m}}+b^{\frac{1}{n}}$

where $\cfrac{1}{m}+\cfrac{1}{n}+\cfrac{1}{k}\lt 1$

Don't panic, it's only maths.  

Are $\frac{1}{m}$, $\frac{1}{n}$ and $\frac{1}{k}$ confined strictly within the space of a one eighth of a sphere of radius $r=\frac{2}{\pi}$ (surface of sphere not included)? Yes. As required.

This will have finite numbers of triplets, $(a^m,\,b^n,\,c^k)$ but their values can be high.

Does $a$, $b$ and $c$ being co-prime make them (their axes) orthogonal here?  As illustrated, positive and orthogonal.

$\{\frac{1}{k},\,c\}$, is defined on the plane extended by $(\frac{1}{n}\otimes b)$ and $(\frac{1}{m}\otimes a)$.  The cross product $(\frac{1}{k}\otimes c)$ lift $\{\frac{1}{k},\,c\}$ into a perpendicular plane.  $(\frac{1}{m}\otimes a)$, $(\frac{1}{n}\otimes b)$ and $(\frac{1}{k}\otimes c)$ forms a rectangular axis, right handed.

What happen to the angle in between with cross product?

It is in $\cfrac{1}{n}$, etc in actual $\cfrac{1}{n}$ should be represented by a vector,

$\cfrac{2}{\pi}e^{i\theta}$, where $\theta=\cfrac{1}{n}\cfrac{\pi}{2}$   the arc length $\cfrac{1}{n}$ divided by the radius, $\cfrac{2}{\pi}$.  This radius constant is in all $a$, $b$ and $c$ and cancels eventually in the equation,

$a^m+b^n=c^k$

Consider, $$y=e^{i\theta}$$

$$ln(y)=i\cfrac{1}{n}\cfrac{\pi}{2}$$

$$-i\cfrac{2ln(y)ln(a)}{\pi}=\frac{1}{n}ln(a)$$

$$y^{-i2ln(a)/\pi}=a^{\frac{1}{n}}$$

$$e^{i\theta-i2ln(a)/\pi}=a^{\frac{1}{n}}$$

$$e^{i\theta}a^{-i2/\pi}=a^{\frac{1}{n}}$$    This is wrong.

$$e^{i\theta}=a^{\frac{1}{n}+i2/\pi}$$

$$Re[e^{i\theta}]=cos(\theta)=a^{\frac{1}{n}}$$

This is where $cos(\theta)$ for the cross product is hiding.

This means the illustration is consistent.  Fermat–Catalan Conjecture, on hold.