[過去ﾛｸﾞ] 高木くんがアクセプトされるまで見守るスレ ★4 (1002ﾚｽ)
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748(1): ◆pObFevaelafK [sage] 2023/07/29(土)20:04 ID:3mRZQXR+(3/7)
Another proof of Legendre's conjecture
Let p_n be the largest prime number among the primes smaller than m^2.
We assume that the following inequalities hold.
p_n<m^2<(m+1)^2<p_(n+1)
According to Cramer's conjecture,
p_(n+1)-p_n<log(p_n)
holds for n≧5.
The following equality must hold when this inequalities hold.
2m+1<log(p_n)
√p_n<m
2√(p_n)+1<2m+1
Since according to Dusart's inequality,
p_n>n(log(n)+log(log(n))-1) holds for n≧2,
2√(n(log(n)+log(log(n))-1))+1<2m+1
holds.
Since according to Dusart's inequality,
p_n<n(log(n)+log(log(n))) holds for n≧6,
log(p_n)<log(n(log(n)+log(log(n))))
holds.
From the above,
2√(n(log(n)+log(log(n))-1))+1<log(n(log(n)+log(log(n))))
holds. However, this equality does not hold for n≧3.
Therefore, there are at least one prime between m^2 and (m+1)^2 for m≧3
since 2m+1<log(p_n) does not hold for n≧6 and the assumption is false.
749(1): ◆pObFevaelafK [sage] 2023/07/29(土)20:09 ID:3mRZQXR+(4/7)
>>748 推敲版
Another proof of Legendre's conjecture
Let p_n be the largest prime number among the primes smaller than m^2.
We assume that the following inequalities hold.
p_n<m^2<(m+1)^2<p_(n+1)
According to Cramer's conjecture,
p_(n+1)-p_n<log(p_n)
holds for n≧5.
The following inequality must hold when this inequalities hold.
2m+1<log(p_n)
√p_n<m
2√(p_n)+1<2m+1
Since according to Dusart's inequality,
p_n>n(log(n)+log(log(n))-1) holds for n≧2,
2√(n(log(n)+log(log(n))-1))+1<2m+1
holds.
Since according to Dusart's inequality,
p_n<n(log(n)+log(log(n))) holds for n≧6,
log(p_n)<log(n(log(n)+log(log(n))))
holds.
From the above,
2√(n(log(n)+log(log(n))-1))+1<log(n(log(n)+log(log(n))))
holds. However, this inequality does not hold for n≧3.
Therefore, there are at least one prime between m^2 and (m+1)^2 for m≧3
since 2m+1<log(p_n) does not hold for n≧6 and the assumption is false.
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