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
9 months 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 - 9 months ago

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

Any ideas?

Steven Jim - 9 months ago

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

Stephen Mellor - 9 months ago

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@Stephen Mellor Anything actually come to your mind yet?

Steven Jim - 9 months 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 - 9 months 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 - 7 months, 4 weeks ago

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Thanks! I’ll check it out.

Steven Jim - 9 months ago

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It might be easier to think about whether the sum can ever be rational for \(k > 1\). Consider the pair of irrational numbers, \(I,J \in \mathbb{R} \backslash \{\mathbb{Q}\}\).

Then \(I+J\) is rational if and only if \(I = R-J\), where \(R \in \mathbb{Q}\).

Now the square-root of any integer is irrational if at least one of the prime factors of that integer has odd multiplicity: in other words, \(\displaystyle \sqrt{n} = \sqrt{p_1^{a_1}\cdot p_2^{a_2} \cdots p_j^{a_j}}\), where \(j,n \in \mathbb{Z}\), and we have \(j\) primes with \(a_i\) multiplicity (\(1 \le i \le j\)). In fact, the \(k^{\text{th}}\) root of any integer is irrational provided the multiplicity of at least one of the prime factors of that number is not a multiple of \(k\): \[\displaystyle \sqrt[k]{n} = \sqrt[k]{p_1^{a_1}\cdot p_2^{a_2}\cdots p_i^{a_j}},\] where at least one \(a_i, \space (1\le i \le j)\), is not a multiple of \(k\).

Now the \(k^{\text{th}}\) root of any prime number is irrational for integers \(k>1\), so we know that for \(k>1, \space \sqrt[k]{2} \notin \mathbb{Q}\). Without even considering whether the other numbers in this series are rational or not, we know that for \(k,n>1, \space \sqrt[k]{n} \neq R-\sqrt[k]{2}\) for some rational number \(R\), so this sum cannot ever be rational (let alone an integer!)


This 'proof' definitely makes a few assumptions like the irrationality of the \(k^{\text{th}}\) roots of prime numbers \(\left(k>1 \in \mathbb{Z}\right)\), or the fact that for \(k,n>1, \space \sqrt[k]{n} \neq R-\sqrt[k]{p}\) for some rational number \(R\), and prime \(p\), but I suppose you could try and prove those individually if you're looking for a rigorous proof.

Akeel Howell - 1 month, 3 weeks ago

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this is for series i think

Arkodeep Paul - 9 months ago

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