# 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 years, 1 month 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

- 3 years, 1 month ago

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

Any ideas?

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Anything actually come to your mind yet?

- 3 years, 1 month 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.

- 3 years, 1 month ago

Thanks! I’ll check it out.

- 3 years, 1 month 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.

- 2 years, 12 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.

- 2 years, 5 months ago

this is for series i think

- 3 years, 1 month ago