Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
@Shashi: Did you mean to write "the square root of $x$ is $y$ such that $y^2=x$"? –  Matt N. Feb 8 '11 at 10:36
    
@Matt: Sorry for typo mistake.. –  Shashi Feb 8 '11 at 10:40
12  
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. –  Jon Feb 8 '11 at 10:46
2  
"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 –  JDH Feb 8 '11 at 12:27
    
The "logic" tag seems unappropiate. –  Bruno Stonek Feb 8 '11 at 22:06
show 3 more comments

2 Answers 2

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 $0$.

share|improve this answer
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.