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750(3): ◆pObFevaelafK [sage] 2023/07/29(土)20:17 ID:3mRZQXR+(5/7)
>>749 再推敲
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))^2
holds for n≧5.

The following inequality must hold when this inequalities hold.
2m+1<(log(p_n))^2

√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)))))^2
holds. However, it is confirmed that this inequality does not hold for
n≧58 by numeric computation.

Therefore, there are at least one prime between m^2 and (m+1)^2 for m≧16
since 2m+1<log(p_n) does not hold for n≧58 and the assumption is false.

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751: ◆pObFevaelafK [sage] 2023/07/29(土)20:18 ID:3mRZQXR+(6/7)
>>750 訂正
×this
〇these

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752: ◆pObFevaelafK [sage] 2023/07/29(土)21:17 ID:3mRZQXR+(7/7)
>>750 訂正
×since 2m+1<log(p_n)
〇since 2m+1<(log(p_n))^2

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753: ◆pObFevaelafK [sage] 2023/07/30(日)09:14 ID:usyTwEio(1/2)
>>750 訂正
×there are
〇there is

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