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Suppose that $ \; \large{(n^{2}+7)^{-n}} \; $ is an odd number.
Now, how can I prove that $ \; \large{n^{2}+4} \; $ is always a positive and even number? Could you show me that, please?

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So is $n$ negative here? –  Thomas Apr 13 '12 at 15:06
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I really don't think (even-or-odd) is a tag that should be used :)... if nothing else, not for this problem. –  Tyler Apr 13 '12 at 15:12
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I'm not sure either! haha –  Tyler Apr 13 '12 at 15:15
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@KerimAtasoy: Ok, but I assume that we are only calling integers even or odd? –  Thomas Apr 13 '12 at 15:20
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I.e. saying that $n^2 + 4$ is even means that $n$ is an integer or maybe a square root. –  Thomas Apr 13 '12 at 15:21
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1 Answer

up vote 4 down vote accepted

I guess that $n$ is a negative integer here, so that exponent $-n$ is positive. To justify this: unless you have extended the usual definition of odd and even beyond integers, saying that $n^2 + 4$ is even means that $n^2$ is an integer. So $n$ is an integer, or $n$ is a square root of an integer. In any case, $n^2 +7$ is also an integer, and so if $(n^2 + 7)^{-n}$ is an integer, then $n$ is necessarily an integer.

Proof by contradiction: Assume that $n^2 + 4$ is odd. Then $n^2$ is odd. Then $n^2 + 7$ is even, and so $(n^2+7)^{-n}$ is even.

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