Q:

The validity of the proofs of the Pythagorean Theorem and the concept of area

this might be a very elemental question but it has been bothering me for a while. Must of the proofs I've seen of the Pythagorean Theorem involve showing that the areas of the squares with side length $a$ and $b$ add up to the area of the square with side length $c$. This is generally done by rearranging triangles. 
My problem with this type of proofs is that they only show that the areas must be the same but don't show that $a^2+b^2=c^2$.
Why must the area of a square with side $a$ be defined as $a^2$. Say for example that you had another way of measuring the surface of a square with a given side length (and it behaves as we would intuitively want area to behave). If this function is called $A$ then the visual proofs of the theorem would only show that $A(a)+A(b)=A(c)$. 
So, does this type of proof works because we just happen to define area as we do, or does $A(a)+A(b)=A(c)$ must imply $a^2+b^2=c^2$? 
Now, if $A(a)+A(b)=A(c)$ does imply $a^2+b^2=c^2$ that would mean that our function $A$ (which behaves as area does) must include the square of the side in its formula. For example $A(x)=kx^2, k>0$ (which does imply the pythagorean theorem). Are there other ways to define the surface of a square such that it behaves as it physically does? Would the visual proofs still be valid?
Thank you!

A:

Any reasonable definition of areas will make the area of a square of side $s$ proportional to $s^2$. I'll give the proof and indicate [in brackets] what "reasonableness" is required to make the proof work. 
Suppose $A$ is a reasonable definition of areas; $A(s)$ is the area of a square of side $s$. [This already presupposes one aspect of "reasonableness", namely that all squares with the same side length have the same area.] For any positive integer $q$, we can partition a square of side $1$ into $q^2$ smaller squares of side $1/q$, just by drawing perpendicular grid lines, at $1/q$ spacing, parallel to the sides of the original square. So each of these smaller squares has area $A(1)/(q^2)$. [Here I'm assuming another aspect of reasonableness: If several squares overlap only along edges, then the area of their union is the sum of their areas.] So $A(1/q)=A(1)(1/q^2)$. Now let $p$ be another positive integer and consider a square of side $p/q$. It can be split by grid lines into $p^2$ squares of side $1/q$. So, by the same reasonableness properties already used above, we get 
$$
A(p/q)=p^2A(1/q)=p^2A(1)(1/q^2)=A(1)(p/q)^2.
$$
So we have $A(x)=A(1)x^2$ for all positive rational $x$. 
As a final reasonableness property, I'll assume that $A$ is a monotone function: if $x<y$ then $A(x)\leq A(y)$. Then, for any positive real number $r$, we have that $A(r)$ is sandwiched between $A(x)$ and $A(y)$ for all positive rational $x$ and $y$ with $x<r<y$. That is, for all such positive rational $x$ and $y$, we have $A(1)x^2\leq A(r)\leq A(1)y^2$. The only number $A(r)$ that satisfies these inequalities, for all such $x$ and $y$, is $A(r)=A(1)r^2$. 
So, up to a constant factor $A(1)$ which amounts to a choice of units, the usual notion of area is the only reasonable one, provided you accept that a reasonable notion of area must have the properties I stated during the proof.
(The last of the reasonableness assumptions could be replaced by assuming that $A$ is continuous. The required conclusion would still follow. I prefer to use monotonicity, because it's easier to state and to understand. In particular, I think my whole argument would have made good sense to Euclid; indeed, for all I know, it might be in Euclid's "Elements".)

