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 $y^2=x$"? – Rudy the Reindeer Feb 08 '11 at 10:36
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@Matt: Sorry for typo mistake.. – Batakj Feb 08 '11 at 10:40
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12Bertrand'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. – Alf Feb 08 '11 at 10:46
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2"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_postulate – JDH Feb 08 '11 at 12:27
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The "logic" tag seems unappropiate. – Bruno Stonek Feb 08 '11 at 22:06
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I removed the "logic" tag. – Matthew Conroy Feb 08 '11 at 22:19
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I was going to say this is a dupe, but then I saw the date of posting... – Aryabhata Jun 09 '11 at 22:26
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@JDH, the verse is about, but not by, Erdos. The author was Nathan Fine. – Gerry Myerson Dec 04 '11 at 06:07
2 Answers
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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 $0$.
Srivatsan
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Ross Millikan
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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).
Fixee
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