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I am not able to give a proof of the following statement: given an integer number $k$, we consider the following expression: $$x=\sqrt{k^3}-\sqrt[3]{k^2}$$ Show that you can get infinite prime numbers from this formula. I tried for $1\le k \le 1e7$ and I found only one prime number: $531457$. Is it possible to show we can get only a finite set of primes using the equation defined above? Many thanks.

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What is the source of this problem? – Aryabhata Mar 20 '12 at 17:33
Do you mean to take the floor of the expression? It doesn't appear to be a natural very often-only for sixth powers – Ross Millikan Mar 20 '12 at 17:37
... and when $k>1$ is a sixth power, $x$ is most certainly never a prime. – Jyrki Lahtonen Mar 20 '12 at 17:39
You have to consider only the $x$ integer. – Riccardo.Alestra Mar 20 '12 at 17:43
Then you mean $x=\left\lfloor \sqrt{k^3}-\sqrt[3]{k^2} \right\rfloor$. Unless you want to round to the nearest integer instead of taking the floor. Otherwise, as people have said, $\sqrt{k^3}-\sqrt[3]{k^2}$ is only an integer if $k=b^6$ for some $b$. – Thomas Andrews Mar 20 '12 at 17:54
up vote 2 down vote accepted

One possible interpretation is that an infinite number of primes appear as divisors in the sequence $\lfloor k^{3/2}-k^{2/3} \rfloor$ for $k \in \mathbb{N}$. This follows from the fact that there are approximately $N^{2/3}$ numbers of this form below $N$ and this growth rate cannot be achieved with only finitely many primes as factors.

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