Elementary proof that $24^n$ is the sum of 3 perfect squares

Question

Proof that $24^n$ is the sum of 3 perfect squares for every $n >= 1$ . I found this problem in a math contest for 5th graders http://www.mategl.com/atasamente%20pentru%20site/OLM%202014%20Prahova%20(5-12).rar , so obviously a very very elementary proof is required (definitely not induction, very little algebra). It is easy to figure out that $24^1 = 2^2 + 2^2 + 2^4 $ and $24^2 = 2^6 + 2^8 +2^8$ but I could not see the pattern from here. I also thought about $24^1 = 2^3 + 2^4 $ and $24^2 = 2^6 + 2^9 $ but I can't see a pattern.

Answer 1

Hint: given that you have split $24^1$ and $24^2$ into sums of three squares, and that a square times a square is a square, can you find a way to conclude?

Answer 2

Now if $k$ is even you can let $m=\frac k2-1$ and $$24^k=(24^m)^224^2=(24^m\cdot 2^3)^2+(24^m\cdot 2^4)^2+(24^m\cdot 2^4)^2$$ Similarly for odd $k$ define $m=\frac 12(k-1)$ and $24^k=(24^m)^2\cdot 24$

Answer 3

\begin{align} 24^{2n + 2} &= 24^2 * 24^{2n} = (2^6 + 2^8 + 2^8)* 24^{2n} = (8* 24^n)^2 + (16*24^n)^2 + (16*24^n)^2 \\ 24^{2n+1} &= 24 * 24^{2n} = ( 2^2 + 2^2 + 2^4 ) * 24^{2n} = (2*24^n)^2 + (2*24^n)^2 + (4*24^n)^2\\ \end{align}

P.S: Thus the OP was the right track, if they would have computed it for $24^3$ and $24^4$, they would have been able to observe a pattern.