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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.

Riccardo.Alestra
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  • What is the source of this problem? – Aryabhata Mar 20 '12 at 17:33
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    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
  • If $k$ is a sixth power of $b$, $x = b^9-b^4 =b^4(b^5-1)$ is not prime. – Desiato Mar 20 '12 at 17:51
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    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
  • If $k=n^6$ and $n>1$, then $$ \sqrt{k^3}-\root 3\of{k^2}=n^9-n^4=n^4(n^5-1)$$ is manifestly not prime.

    And if $k$ is not a sixth power, then the two terms are non-rational algebraic integers of degrees two and three respectively, so their sum cannot be a rational integer.

    Something doesn't add up. Which value of $k$ gives $$531457=3^{12}+2^4?$$

     – Jyrki Lahtonen Mar 20 '12 at 17:56
  • To get $531457$, you need to start from $k=6564.01672376\ldots$. If you start from $k=6564$ then you get $531454.968\ldots$. – Henry Mar 20 '12 at 18:26
  • You should not write "infinite prime numbers" if you mean "infinitely many prime numbers". The phrase "infinite prime numbers", used correctly, means "prime numbers, each one of which, by itself, is infinite". – Michael Hardy Mar 20 '12 at 18:32

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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.

WimC
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