How to prove that $x^2+y^2+z^2=14^n$ holds for distinct integers $x,y,z$ for every natural $n$? Prove that for all natural numbers $n$, there exist distinct integers $x, y, z$ for which, 

$x^2+y^2+z^2=14^n$

How to prove this using mathematical induction?

Some context: 
A related question asks for solutions of $x^2 + y^2 + z^2 = 3^{10}$ where the asker first tries to express $3^1$ and $3^2$ as sums of three squares and then combine these to construct a representation of $3^3$, and so on. However, none of the answers seem to use this approach.
Legendre's three square theorem  states that $n \in \mathbb{N}$ can be expressed as a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$, and $14^k$ clearly isn't. However, it does not guarantee that  $x,y,z$ are distinct and applying it  is not (at least not directly) a proof by induction.

For $n=1$ we have
$$14=1^2+2^2+3^2$$
For $n=2$ we have
$$14^2=196=4^2+6^2+12^2$$
If $14^n=x^2+y^2+z^2$ then
$$14^{n+2}=14^2(x^2+y^2+z^2)=(14x)^2+(14y)^2+(14z)^2$$
so you can do induction on even and odd numbers separately.
