Question

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

Answer 1

Hint: Don't think too far on this one, it's a fifth grade question. The answer is 2-3 lines long.

Answer 2

(Induction)

Base case : \(10^1 = 3^2+1^2\)

Induction step :
\[10^{n+1} = 10*10^n = (3^2+1^2)(a^2 + b^2) = (3a-b)^2 + (3b+a)^2\]

Answer 3

used brahmagupta's identity
http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

Answer 4

Oh, that's a good one! Haven't thought of it.

Answer 5

please don't put the 5th grade solution yet, im still trying...

Answer 6

Oh, alright - I'm not going to spoil the fun then.

Answer 7

\[\Large\sf{
10=1^2+3^2\\
100=6^2+8^2\\\\
\\
n~odd~\\n=2m+1\\
10^n=(10^m)^2+(3\times10^m)^2\\
n~even\\
n=2m\\
10^n=(10^{m-1}\times6)^2+(10^{m-1}\times8)^2
}\]

Answer 8

I don't know this is a fifth grade solution....
In our country this would be 8th grade....

Answer 9

thats very clevever!

Answer 10

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

Answer 11

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.

Answer 12

Damn it, the other way around.

Answer 13

I found it in a 5'th grade textbook for the math olympiad.

Answer 14

please post such qs.. I like them....

Answer 15

I'm glad you do, I really like them too! Let me look for another one...

Answer 16

very clever solution...