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.