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<record version="1" id="8840">
 <title>Legendre's conjecture</title>
 <name>LegendresConjecture</name>
 <created>2007-01-29 18:07:57</created>
 <modified>2007-01-29 18:07:57</modified>
 <type>Conjecture</type>
 <creator id="13766" name="PrimeFan"/>
 <author id="13766" name="PrimeFan"/>
 <classification>
	<category scheme="msc" code="11A41"/>
 </classification>
 <related>
	<object name="BrocardsConjecture"/>
 </related>
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 <content>(Adrien-Marie Legendre) There is always a prime number between a square number and the next. To put it algebraically, given an integer $n &gt; 0$, there is always a prime $p$ such that $n^2 &lt; p &lt; (n + 1)^2$. Put yet another way, $(\pi((n + 1)^2) - \pi(n^2)) &gt; 0$, where $\pi(x)$ is the prime counting function.

This conjecture was considered unprovable when it was listed in Landau's problems in 1912. Almost a hundred years later, the conjecture remains unproven even as similar conjectures (such as Bertrand's postulate) have been proven.

But progress has been made. Chen Jingrun proved a slightly weaker version of the conjecture: there is either a prime $n^2 &lt; p &lt; (n + 1)^2$ or a semiprime $n^2 &lt; pq &lt; (n + 1)^2$ (where $q$ is a prime unequal to $p$). Thanks to computers, brute force searches have shown that the conjecture holds true as high as $n = 10^5$. % I look forward to updates. Mathematica confirms to ten thousand in 17 seconds, I'm still waiting for an answer on one million</content>
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