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Prove that a perfect square is either a multiple of $4$ or of the form $4q+1$ for some $q\in \mathbb{Z}$.

Any ideas on how to start? Do I use a proof by contraposition? Also what's the definition of a perfect square?

Michael Albanese
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1 Answers1

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Hint: A perfect square is an integer $k$ such that $k = n^2$ for some integer $n$. It follows that $k$ must be non-negative and that $n$ can be chosen to be non-negative. As for approaching the problem, first break it up into two cases: $k$ even and $k$ odd.

Michael Albanese
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  • Okay. But then how do I relate k even or odd with 4 or 4q+1? Do I write k = 2q for even and k = 2q+1 for odd? –  Oct 08 '13 at 13:02
  • I'd first write $k = n^2$, using the fact that $k$ is a perfect square. Then, the parity of $k$ (i.e. whether $k$ is even or odd) is related to the parity of $n$. Then I would write $n$ as $2q$ or $2q+1$, depending on the parity of $n$. – Michael Albanese Oct 08 '13 at 13:04
  • This is what I have so far: $k = n^2$ as k is a perfect square. Case 1: n is even so $n = 2q$. Then $k = (2q)^2 = 4q^2$.

    Case 2: n is odd so $n = 2q+1$. Then $k = (2q+1)^2 = 4q(q+1)+1$.

    Is this okay?

     –  Oct 08 '13 at 13:11
  • Looks good to me, but be careful. The $q$ you have used is different to the $q$ mentioned in the question. I would use a different letter, say $p$, so that if $n = 2p+1$, $k = n^2 = 4p(p+1) + 1 = 4q+1$, where $q = p(p+1)$. – Michael Albanese Oct 08 '13 at 13:16
  • Oh I see. But for case 1, how do I make $p^2$ disappear? Do I let $q = 1 = p^2$? –  Oct 08 '13 at 13:23
  • Yes, let $q = p^2$. I don't know why you set it equal to $1$. – Michael Albanese Oct 08 '13 at 13:25
  • Okay. Thanks a lot for your help! –  Oct 08 '13 at 13:25