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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?

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    @Shashi: Did you mean to write "the square root of $x$ is $y$ such that $y^2=x$"?2011-02-08
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    @Matt: Sorry for typo mistake..2011-02-08
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    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.2011-02-08
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    "Chebyshev said it before, and I say it again, there is always a prime beteween $n$ and $2n$." -Erdos. http://en.wikipedia.org/wiki/Bertrand's_postulate2011-02-08
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    The "logic" tag seems unappropiate.2011-02-08
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    I removed the "logic" tag.2011-02-08
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    I was going to say this is a dupe, but then I saw the date of posting...2011-06-09
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    @JDH, the verse is *about*, but not *by*, Erdos. The author was Nathan Fine.2011-12-04

2 Answers 2

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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 $0$.

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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).