Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares. Show that $\forall n\in\mathbb{N}$, $14^n$ can be represented as a sum of three perfect squares.
I checked $(\mod 7)$ and deduced that the three squares can be $1,4,2(\mod 7) $ or all divisible by $7$ and for $n \ge 2$ they must be all divisible by $4$. But I am stuck here.
 A: When $n=2k$ is even you are done since $$14^n=\left(14^k\right)^2+0^2+0^2$$ and when $n=2k+1$ is odd we can write $$14^n=14^{2k}\cdot(9+4+1)=\left(3\cdot 14^{k}\right)^2+\left(2\cdot 14^{k}\right)^2+\left(14^{k}\right)^2.$$
A: It is known that an integer is a sum of three squares if and only if the integer is not of the form $4^a(8b+7)$ so we show that all powers of $14$ are not of this form.
$$14^n=(2(8-1))^n=2^n(8-1)^n=2^n(8M\pm1)$$
If $n$ is even we have $$14^n=4^k(8M+1)$$
If $n$ is odd we have $$14^n=2\cdot4^k(8M-1)=4^k(8N-2)=4^k(8T+6)$$ In both cases, when $n$ is even and odd, we see that $14^n$ is not of the form $4^a(8b+7)$ hence it is sum of three squares.
A: You can use induction.
Step 1:
When $n = 1$, we have $14 = 3^2 + 2^2 + 1^2$
When $n = 2$, we have $14^2 = 12^2 + 6^2 + 4^2$
Step 2: Suppose the statment is right for all $1 \le k \le n$. 
In particular, $\exists a, b, c > 0$ s.t. $14^{n-1} = a^2 + b^2 + c^2$
Then for $n + 1$,
$14^{n+1} = 14^{n-1+2} = 14^{n-1} \times 14^2 $
$= 14^2\times a^2 + 14^2\times b^2 + 14^2\times c^2$
$= (14 \times a)^2 + (14 \times b)^2 + (14 \times c)^2$
Finally we can say the statement is true.
