Consider an equation from the Fermat's Last Theorem without the integer condition (I'll call it FLT equation) $$x^p+y^p=z^p,\tag{1}$$ where $x$, $y$ and $z$ are nonnegative real numbers and $p\ge2$. (FLT states that if $p$ is an integer such that $p\ge3$, then the above equation has no integer valued solution. But, this is not the point of my question.)
If $p=2$, then the equation (1) is closely related to the concept of right angle. That is, if three points $X$, $Y$ and $Z$ are vertices of a triangle in $\mathbb R^2$ and $x$, $y$ and $z$ are the lengthes of the corresponding segments (i.e. $x=\overline{YZ},\quad y=\overline{ZX},\quad z=\overline{XY}\:$), then
$\triangle XYZ$ is a right triangle$\left(\angle Z=\frac\pi2\right)$ if and only if the equation (1) holds.
This is just the Pytagorean Theorem and its converse.
Can we make a similar statement when $p>2$?
A modification of (1) $$\sqrt[p]{x^p+y^p}=z,\tag{2}$$ is somewhat similar to the definition of Lp norm. So, I want say like
$\triangle XYZ$ is the right triangle$\left(\angle Z=\frac\pi2\right)$ if and only if the equation (1) holds.
in the Lp sense.
Here is my opinion. If $p>2$, then the Lp space (the set $\mathbb R^2$ with the L-p norm) is a normed space. But it is not an inner product space($l^p$ space not having inner product). So it is not possible to define the notion of angle, where the formal definition of angle that I learned is as follow (Here $\theta$ is an angle between vectors $u$ and $v$ in an inner product space.); $$\cos\theta=\frac{<u,v>}{||u||\cdot||v||}$$
So I think that the above statement that I made is not making any sense at all. In other words, speaking of angle or orthogonality in a normed space without the parallogram law is just nonsense. Am I right? Or is there any ways to relate FLT equation and the orthogonality in the Lp sense? And is the orthogonality always defined by using the concept of inner product? ($u\perp v\iff <u,v> =0$)
