Legendre’s conjecture

(Adrien-Marie Legendre) There is always a prime number between a square number and the next. To put it algebraically, given an integer $n>0$ , there is always a prime $p$ such that $n^{2}<p<(n+1)^{2}$ . Put yet another way, $(\\pi((n+1)^{2})-\\pi(n^{2}))>0$ , 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 $n^{2}<p<(n+1)^{2}$ or a semiprime $n^{2}<pq<(n+1)^{2}$ (where $q$ is a prime unequal to $p$ ). Thanks to computers, brute force searches have shown that the conjecture holds true as high as $n=10^{5}$ .

单词	LegendresConjecture
释义	Legendre’s conjecture (Adrien-Marie Legendre) There is always a prime number between a square number and the next. To put it algebraically, given an integer $n>0$ , there is always a prime $p$ such that $n^{2}<p<(n+1)^{2}$ . Put yet another way, $(\\pi((n+1)^{2})-\\pi(n^{2}))>0$ , 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 $n^{2}<p<(n+1)^{2}$ or a semiprime $n^{2}<pq<(n+1)^{2}$ (where $q$ is a prime unequal to $p$ ). Thanks to computers, brute force searches have shown that the conjecture holds true as high as $n=10^{5}$ .
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