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anonymous
 one year ago
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
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hint: Don't think too far on this one, it's a fifth grade question. The answer is 23 lines long.

rational
 one year ago
Best ResponseYou've already chosen the best response.2(Induction) Base case : \(10^1 = 3^2+1^2\) Induction step : \[10^{n+1} = 10*10^n = (3^2+1^2)(a^2 + b^2) = (3ab)^2 + (3b+a)^2\]

rational
 one year ago
Best ResponseYou've already chosen the best response.2used brahmagupta's identity http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, that's a good one! Haven't thought of it.

rational
 one year ago
Best ResponseYou've already chosen the best response.2please don't put the 5th grade solution yet, im still trying...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, alright  I'm not going to spoil the fun then.

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\sf{ 10=1^2+3^2\\ 100=6^2+8^2\\\\ \\ n~odd~\\n=2m+1\\ 10^n=(10^m)^2+(3\times10^m)^2\\ n~even\\ n=2m\\ 10^n=(10^{m1}\times6)^2+(10^{m1}\times8)^2 }\]

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2I don't know this is a fifth grade solution.... In our country this would be 8th grade....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well done! For n=even, n=2k we have 10^n=10^(2k2)*10^2=(10^(k1)*6)^2 + (10^(k1)*8)^2 Similar for n=odd,n=2k+1 we have 10^n=10^(2k+1)=(10^k*3)^2 + (10^k)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Damn it, the other way around.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found it in a 5'th grade textbook for the math olympiad.

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2please post such qs.. I like them....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm glad you do, I really like them too! Let me look for another one...

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0very clever solution...
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