I need require assistance in proving that $$2^{n} s$$ can be written as the sum of two perfect squares, where $$s$$ is a sum of two perfect squares. My teacher has told me I'm not allowed to use induction and I'm not really sure how to go about it.

If anyone could nudge me in the night direction, I'd be appreciative.

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    $\begingroup$ Hint: Think about the case when n is even and the case when n is odd $\endgroup$ – Amr May 28 '13 at 23:54
  • $\begingroup$ Relevant question (possible duplicate): math.stackexchange.com/q/100506/264 $\endgroup$ – Zev Chonoles May 28 '13 at 23:55
  • $\begingroup$ I've thought about that...but I get that it can be expressed as the sum of 2 perfect squares when 'n' is odd and it can't when 'n' is even but I'm required to show it works for all natural n :( $\endgroup$ – user78416 May 28 '13 at 23:55
  • $\begingroup$ As a nuclear overkill, you can use Fermat's classification of numbers which can be written as the sum of 2 squares. But please just follow Amr's hint. $\endgroup$ – Calvin Lin May 28 '13 at 23:56
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    $\begingroup$ Because you say you have trouble when $n$ is even, I thought it might help to remind you that $0$ is a perfect square. E.g., $2^2=2^2+0^2$. $\endgroup$ – Jonas Meyer May 28 '13 at 23:57

In the case that $ n $ is odd, let $ s = k^2 $ and $ n = 2m + 1$. Hence $$ 2^{2m + 1}k^2 = \left(2^m k\right)^2 + \left(2^mk\right)^2 $$ Otherwise, $ n = 2m $ and we have that $$ 2^{2m}k^2 = \left(2^m k \right)^2 + 0^2 $$ No induction needed here. $ m $ and $ k $ are both in $ \mathbb{N} \cup \{0\} $.

You might think the zero is "cheating" but it is definitely necessary in some cases such as $ 2^n s = 16 $.

  • $\begingroup$ There is a problem when n is odd $\endgroup$ – Amr May 29 '13 at 0:03
  • $\begingroup$ Thank you. I have fixed it. $\endgroup$ – Jon Claus May 29 '13 at 0:06
  • $\begingroup$ I'm sorry, I had a typo in the original problem :( s is not a perfect square, it is the sum of two perfect squares, sorry :( $\endgroup$ – user78416 May 29 '13 at 0:43

Hint: If the numbers $x$ and $y$ can each be written as the sum of 2 perfect squares, then so can the number $xy$. This is known as the Brahmagupta-Fibonacci Identity.

You can either prove this statement yourself, or click on the Wikipedia link.

  • $\begingroup$ So that covers the case of even n! Thanks very much! :D $\endgroup$ – user78416 May 29 '13 at 1:55

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