Intuition tells no, but how do you do this?!

For an integer \(a>1\), can \(\sum _{ n=1 }^{ k }{ \sqrt [ a ]{ n } } \) ever be an integer for any integer \(k>1\)?

For \(a=2\), the answer is most likely no. However, I don't have one single clue of how to prove this with \(a=2\), let alone the generalization!

Is there a definitive answer for all integer \(a\)? If so, how?

(It might have something to do with Calculus here, it looks like some kind of series...)

Note by Steven Jim
3 months, 1 week ago

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It doesn't make sense to me, do you mean the n to be an x? Otherwise, you are just multiplying the root by the value of k

Stephen Mellor - 3 months, 1 week ago

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Thanks for your info. I didn't have much time when I was typing down.

Any ideas?

Steven Jim - 3 months, 1 week ago

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I agree that it is probably no, but not sure how to prove it

Stephen Mellor - 3 months, 1 week ago

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@Stephen Mellor Anything actually come to your mind yet?

Steven Jim - 3 months, 1 week ago

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While doing some research about this, I came across this blog post from almost a decade ago, which applies to your problem for \(a = 2.\) As for higher values of \(a,\) I haven't found anything yet, but perhaps the methods in the blog post can help.

Steven Yuan - 3 months, 1 week ago

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Yes I believe this result implies the problem for \(a=2\) can never yield an integer. If your sum is \(\sum_{k=1}^n \sqrt{k} =N \), then setting \(s_1=1-N, r_1=1,\) and \(k=r_ks_k^2\) for squarefree \(r_k\), \(2 \leq k \leq n\), then \(\sum_{k=1}^n s_k \sqrt{r_k} =0 \), contradicting the result in the blog.

Jason Martin - 2 months ago

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Thanks! I’ll check it out.

Steven Jim - 3 months, 1 week ago

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this is for series i think

Arkodeep Paul - 3 months, 1 week ago

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