Find the values of $x \in \mathbb{Z}$ such that there is no prime number between $x$ and $x^2$. Is there any such number?

  • $\begingroup$ @Shashi: Did you mean to write "the square root of $x$ is $y$ such that $y^2=x$"? $\endgroup$ – Rudy the Reindeer Feb 8 '11 at 10:36
  • $\begingroup$ @Matt: Sorry for typo mistake.. $\endgroup$ – Shashi Feb 8 '11 at 10:40
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    $\begingroup$ Bertrand's postulate guarantees the existence of a prime between $n$ and $2n$ for all integers $n > 1$. Therefore there are no non-trivial examples of the phenomenon you describe. $\endgroup$ – Jon Feb 8 '11 at 10:46
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    $\begingroup$ "Chebyshev said it before, and I say it again, there is always a prime beteween $n$ and $2n$." -Erdos. en.wikipedia.org/wiki/Bertrand's_postulate $\endgroup$ – JDH Feb 8 '11 at 12:27
  • $\begingroup$ The "logic" tag seems unappropiate. $\endgroup$ – Bruno Stonek Feb 8 '11 at 22:06

Despite the comments about Bertrand's postulate, there is still the range $-\sqrt{2} \le x \le \sqrt{2}$. If you want $x$ a natural number, there is $1$ and maybe $0$.


Given the current wording of the question, you can set $x$ to any integer in $\{-1, 0, 1\}$ and there will be no prime between $x$ and $x^2$. For any other integer $x$, there will always be a prime between $x$ and $x^2$ (as noted in the comments to your question).


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