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

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

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

Is there a definitive answer for all integer aa? If so, how?

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

Note by Steven Jim
1 year, 7 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 - 1 year, 7 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 - 1 year, 7 months ago

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

Stephen Mellor - 1 year, 7 months ago

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@Stephen Mellor Anything actually come to your mind yet?

Steven Jim - 1 year, 7 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.a = 2. As for higher values of a,a, I haven't found anything yet, but perhaps the methods in the blog post can help.

Steven Yuan - 1 year, 7 months ago

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Thanks! I’ll check it out.

Steven Jim - 1 year, 7 months ago

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Yes I believe this result implies the problem for a=2a=2 can never yield an integer. If your sum is k=1nk=N\sum_{k=1}^n \sqrt{k} =N , then setting s1=1N,r1=1,s_1=1-N, r_1=1, and k=rksk2k=r_ks_k^2 for squarefree rkr_k, 2kn2 \leq k \leq n, then k=1nskrk=0\sum_{k=1}^n s_k \sqrt{r_k} =0 , contradicting the result in the blog.

Jason Martin - 1 year, 6 months ago

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

Then I+JI+J is rational if and only if I=RJI = R-J, where RQR \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, n=p1a1p2a2pjaj\displaystyle \sqrt{n} = \sqrt{p_1^{a_1}\cdot p_2^{a_2} \cdots p_j^{a_j}}, where j,nZj,n \in \mathbb{Z}, and we have jj primes with aia_i multiplicity (1ij1 \le i \le j). In fact, the kthk^{\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 kk: nk=p1a1p2a2piajk,\displaystyle \sqrt[k]{n} = \sqrt[k]{p_1^{a_1}\cdot p_2^{a_2}\cdots p_i^{a_j}}, where at least one ai, (1ij)a_i, \space (1\le i \le j), is not a multiple of kk.

Now the kthk^{\text{th}} root of any prime number is irrational for integers k>1k>1, so we know that for k>1, 2kQk>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, nkR2kk,n>1, \space \sqrt[k]{n} \neq R-\sqrt[k]{2} for some rational number RR, so this sum cannot ever be rational (let alone an integer!)


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

Akeel Howell - 1 year ago

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this is for series i think

Arkodeep Paul - 1 year, 7 months ago

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Bennie George - 1 week, 1 day ago

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How does this relate?

Steven Jim - 1 week, 1 day ago

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