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.
