Challenging 5'th grade question for math enthusiasts! Prove that for any natural number n > 0, 10^n can be written as the sum of two square numbers. For example: 10^1=1^2+3^2 10^2=6^2+8^2 and so forth

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Challenging 5'th grade question for math enthusiasts! Prove that for any natural number n > 0, 10^n can be written as the sum of two square numbers. For example: 10^1=1^2+3^2 10^2=6^2+8^2 and so forth

Mathematics
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Hint: Don't think too far on this one, it's a fifth grade question. The answer is 2-3 lines long.
(Induction) Base case : \(10^1 = 3^2+1^2\) Induction step : \[10^{n+1} = 10*10^n = (3^2+1^2)(a^2 + b^2) = (3a-b)^2 + (3b+a)^2\]
used brahmagupta's identity http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

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Oh, that's a good one! Haven't thought of it.
please don't put the 5th grade solution yet, im still trying...
Oh, alright - I'm not going to spoil the fun then.
\[\Large\sf{ 10=1^2+3^2\\ 100=6^2+8^2\\\\ \\ n~odd~\\n=2m+1\\ 10^n=(10^m)^2+(3\times10^m)^2\\ n~even\\ n=2m\\ 10^n=(10^{m-1}\times6)^2+(10^{m-1}\times8)^2 }\]
I don't know this is a fifth grade solution.... In our country this would be 8th grade....
thats very clevever!
Well done! For n=even, n=2k we have 10^n=10^(2k-2)*10^2=(10^(k-1)*6)^2 + (10^(k-1)*8)^2 Similar for n=odd,n=2k+1 we have 10^n=10^(2k+1)=(10^k*3)^2 + (10^k)^2
Ah, I made a big mess. But yeah, that was the idea - take two cases for n=odd and n=even and replace n with 2k and 2k+1 respectively.
Damn it, the other way around.
I found it in a 5'th grade textbook for the math olympiad.
please post such qs.. I like them....
I'm glad you do, I really like them too! Let me look for another one...
very clever solution...

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