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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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    @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).