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...)

#NumberTheory

Easy Math Editor

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Comments

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 years, 3 months ago

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

Any ideas?

Steven Jim - 3 years, 3 months ago

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

Stephen Mellor - 3 years, 3 months ago

@Stephen Mellor – Anything actually come to your mind yet?

Steven Jim - 3 years, 3 months ago

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 years, 3 months ago

Thanks! I’ll check it out.

Steven Jim - 3 years, 3 months ago

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 - 3 years, 2 months ago

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 - 2 years, 8 months ago

this is for series i think

Arkodeep Paul - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$