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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?2012-04-13
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    @Thomas: Well, the exact value of $ \; n \; $ isn't given particularly in the problem.2012-04-13
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    False: There is some $n\approx-0.552$ with $(n^2+7)^{-n1}=3$, and for this $n$, $n^2$ is not an integer.2012-04-13
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    Ok, I was just thinking if $n$ was positive, say equal to 2, then you would have a fraction $\frac{1}{(n^2 + 7)^{-1}}$ and I don't know what it means for such to be odd or even.2012-04-13
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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.2012-04-13
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    @HaraldHanche-Olsen Ok, I might have misunderstood. I assumed that $n$ was an integer. Kerim, is $n$ assumed to be an integer?2012-04-13
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    @TylerBailey: Sorry but, I wasn't sure about the tag, so... :) You could edit if you'd like to... :)2012-04-13
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    @Thomas: It is probably intended for $n$ to be an integer, but in poking fun at the problem I chose to disregard that.2012-04-13
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    I'm not sure either! haha2012-04-13
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    I think the tag is probably fine. You might add another tag though?2012-04-13
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    @TylerBailey: :D OK, that was funny... :D I'm still thinking over some other possible tags, OK... :)2012-04-13
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    @Thomas: Well, there is nothing given more about that $ \; n \; $ particularly...2012-04-13
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    @KerimAtasoy: Ok, but I assume that we are only calling integers even or odd?2012-04-13
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    I.e. saying that $n^2 + 4$ is even means that $n$ is an integer or maybe a square root.2012-04-13
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    @Thomas: YES, YOU'RE RIGHT!... :D It should be an integer, I think, right...? :)2012-04-13
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    @KerimAtasoy: Yes, I think so. Otherwise the problem does make sense to me.2012-04-13
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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.