初定本
質數分布定理,所以究質數之分佈。
質數之以無窮
假設質數有窮,則有
p 1 , p 2 , ⋯ ⋯ , p n , {\displaystyle p_{1},p_{2},\cdots \cdots ,p_{n},} p即質數(prime number)
令諸質數相乘,復加壹,得r
[ ∏ k = 1 n p k ] + 1 = r {\displaystyle \left[\textstyle \prod _{k=1}^{n}\displaystyle p_{k}\right]+1=r}
令r除以質數,得
r ÷ p j = η + 1 p j , η ∈ Z , 1 p j ∈ ( 0 , 1 2 ] {\displaystyle r\div p_{j}=\eta +{\frac {1}{p_{j}}},\eta \in \mathrm {Z} ,{\frac {1}{p_{j}}}\in (0,{\frac {1}{2}}]}
故 r p j ∉ Z {\displaystyle {\frac {r}{p_{j}}}\not \in \mathrm {Z} } ,故r屬質數,有悖於假設,故題得證。
質數皆相隣於 6 n , n ∈ N ∗ {\displaystyle 6n,n\in \mathrm {N^{*}} } ,且 2 , 3 {\displaystyle 2,3} 除外。
2 n ′ ÷ 2 = n ′ , n ′ ∈ N ∗ {\displaystyle 2n'\div 2=n',n'\in \mathrm {N^{*}} }
故
6 n , 6 n + 2 , 6 n + 4 ∈ 2 n {\displaystyle 6n,6n+2,6n+4\in 2n}
同理
3 n ′ ÷ 3 = n ′ , n ′ ∈ N ∗ {\displaystyle 3n'\div 3=n',n'\in \mathrm {N^{*}} }
6 n + 3 ∈ N ∗ {\displaystyle 6n+3\in \mathrm {N^{*}} }
故命題得證
十進制
2,3,5,7,11,13,17,19,etc
六進制
2,3,5,11,15,21,25,31,etc
質數漸稀
由定理二可知
p 3 = 6 − 1 , p 4 = 6 + 1 , p 5 = 2 × 6 − 1 , p 6 = 2 × 6 + 1 , e t c {\displaystyle p_{3}=6-1,p_{4}=6+1,p_{5}=2\times 6-1,p_{6}=2\times 6+1,\mathrm {etc} }
又
p 3 ⋅ p 3 = ( 6 − 1 ) 2 = 6 2 − 2 ⋅ 6 + 1 , p 3 ⋅ p 4 = 6 2 − 1 , e t c {\displaystyle p_{3}\cdot p_{3}=(6-1)^{2}=6^{2}-2\cdot 6+1,p_{3}\cdot p_{4}=6^{2}-1,\mathrm {etc} }
且謂之質數分布缺陷,且缺陷可無限耦合疊加,故命題得證。