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.
Prime numbers in a formula
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0You 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". – 2012-03-20
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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.