Where does the power $2$ come from in the Pythagorean theorem? So
$$a^2+ b ^2 =c^2$$
in a right triangle, but where does the power $2$ come from?
I know we can use different metrics in the Euclidean space. If we use the $p$-metrics, where $p$ is in place of $2$, the case $p=2$ is the only one that makes $\mathbb{R}^n$ an inner product space (right?). So it's the only case where we can even talk about right triangles.
But I guess my question is then that is there some sort of underlying (physical) reason for the exponent $2$?
 A: As was answered in the comments, the $2$ in the Pythagorean theorem simply stems from the fact that it talks about the areas of squares.
Regarding your other question, you are right. The Euclidean norm is the only $p$-norm that makes $\mathbb{R}^n$ into an inner product space and thus the only norm w.r.t. which it makes sense to speak of right angles (i.e. orthogonality).
You can see this by the following:

Fact: A normed real vector space $(V,\|\cdot\|)$ can be equipped with an inner product $\langle\cdot,\cdot\rangle$ such that $\|x\|^2=\langle x,x\rangle$ if and only if the parallelogram identity holds:
  $$2(\|x\|^2+\|y\|^2)=\|x+y\|^2+\|x-y\|^2$$
  for all $x,y\in V$.

A: For another perspective, look at Bhaskara's proof (http://www.geom.uiuc.edu/~demo5337/Group3/Bhaskara.html).  You inscribe a square in a larger square by properly selecting a point on each edge of the larger square and connecting them in rotational order.  Then you show that the area of the smaller square, whose side is the hypotenuse of a right triangle, is correct for the Pythagorean theorem.
This construction relies on the fact that a square is self-dual. That seems so obvious we tend overlook it -- but self-duality fails for measure polytopes in higher dimensions.  There are too few faces in a cube or hypercube to match up with the vertices, and a Bhaskara-type construction is doomed to fail.  From this point of view the exponent 2 ultimately comes from a unique property of two-dimensional space.
