Solution to For n∈Z, prove n^2 is odd if and only if n is odd. - Sikademy
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For n∈Z, prove n^2 is odd if and only if n is odd.

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For n\in\Z, let us prove that n^2 is odd if and only if n is odd.

Let n be an odd number. Then n=2k+1 for some k\in\Z. Then n^2=(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1=2s+1, where s=2k^2+2k\in\Z. Therefore, n^2 is odd.

Let n^2 be an odd number. Let us prove that n is odd using the method by contradiction. Suppose that n is not odd, and hence n is even, that is n=2k for some k\in\Z. Then n^2=(2k)^2=4k^2=2(2k^2)=2m, where m=2k^2\in\Z. It follows that n^2 is even. This contradiction proves that n is odd.


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