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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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    Well, I did my best for the tag... :) I was working with digits, numbers, natural numbers, positive natural numbers, integers, even numbers, odd numbers, consecutive numbers, etc... :)2012-04-13

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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.