<s>[INST] Elementary proof that $24^n$ is the sum of 3 perfect squares 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.
 [/INST] \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. 
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