anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Hint: Don't think too far on this one, it's a fifth grade question. The answer is 2-3 lines long.
rational
  • rational
(Induction) Base case : \(10^1 = 3^2+1^2\) Induction step : \[10^{n+1} = 10*10^n = (3^2+1^2)(a^2 + b^2) = (3a-b)^2 + (3b+a)^2\]
rational
  • rational
used brahmagupta's identity http://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Oh, that's a good one! Haven't thought of it.
rational
  • rational
please don't put the 5th grade solution yet, im still trying...
anonymous
  • anonymous
Oh, alright - I'm not going to spoil the fun then.
amilapsn
  • amilapsn
\[\Large\sf{ 10=1^2+3^2\\ 100=6^2+8^2\\\\ \\ n~odd~\\n=2m+1\\ 10^n=(10^m)^2+(3\times10^m)^2\\ n~even\\ n=2m\\ 10^n=(10^{m-1}\times6)^2+(10^{m-1}\times8)^2 }\]
amilapsn
  • amilapsn
I don't know this is a fifth grade solution.... In our country this would be 8th grade....
rational
  • rational
thats very clevever!
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
Damn it, the other way around.
anonymous
  • anonymous
I found it in a 5'th grade textbook for the math olympiad.
amilapsn
  • amilapsn
please post such qs.. I like them....
anonymous
  • anonymous
I'm glad you do, I really like them too! Let me look for another one...
welshfella
  • welshfella
very clever solution...

Looking for something else?

Not the answer you are looking for? Search for more explanations.