It is possible to show square as sum n distinc squares, where n > 2? We know that $5^2 = 3^2 + 4^2$.
But it is possible to show $a^2$ as the sum of three or more distinct squares?
Something like that exist?
$a^2 = b^2 + c^2 + d^2$, where $b$, $c$, $d$ are distinct natural numbers.
Thanks a lot!
 A: Legendre's three-square theorem states that every number that is not of the form
$$4^n(8m+7)$$
can be written as a sum of exactly three squares. Since most (all?) squares are not of this form, the answer to your question is Yes.
A: The complete solution for your particular case ($n=3$) was given by Catalan — just about any elementary number theory text will give the general integer parameterization of the primitive solutions, which [assuming $b$ is odd] is
\begin{align}
a &= p^2+q^2+r^2+s^2,  \\
b &= p^2+q^2-r^2-s^2,  \\
c &= 2(ps+qr), \\
d &= 2(pr-qs).
\end{align}
The complete answer for any $n \ge 2$ is given in several papers. A good introduction is Bradley’s paper Equal Sums of Squares. For a more technical matrix-based approach, read Barnett and Mendel’s paper On Equal Sums of Squares.
A: Yes, they do exist! An example is $(2, 3, 6, 7)$. $\sqrt{2^2+3^2+6^2}=7$. These are called Pythagorean quadruples. They exist in the form $(a,b,c,d)$. There is a Wikipedia artible about this: click here. 
A Pythagorean quadruple is mostly used to calculate the space diagonal ($d$) of a cuboid, which is equal to $\sqrt{a^2+b^2+c^2}$, where $|a|$, $|b|$ and $|c|$ are the length, width and height of the cuboid.  Here is an example:

(source: smpn1bontang.org) 
$QW$ is the space diagonal, which is equal to $\sqrt{PS^2+RS^2+WS^2}$
