Greener Journal of
Science, Engineering and Technological Research Vol. 9(1), pp. 67, 2019 ISSN: 22767835 Copyright ©2019, the
copyright of this article is retained by the author(s) DOI Link: http://doi.org/10.15580/GJSETR.2019.1.040919067
http://gjournals.org/GJSETR 

A number of conjectures are likely to
setup
Zhou Mi^{ 1*}; Hong Wei Shi ^{2}
^{1} Suqian Economy and Trade Vocational School
^{2} Suqian College
^{ }
ABSTRACT 



An
unresolved conjecture in number theory has
great possibility to set up. 


ARTICLE INFO 



Article
No.: 040919067 Type: Short Comm DOI: 10.15580/GJSETR.2019.1.040919067 
Submitted: 09/04/2019 Accepted:
13/04/2019 Published: 19/04/2019 
*Corresponding
Author Zhou Mi Email: zhoumi19920626@
163.com 
Keywords:










INTRODUCTION:
Have
an unresolved conjecture in number theory,
Q1=2,
q2=3, q3=5, Q4, Q5,,,, from small to large order write all the prime numbers,
Qk=q1*q2*q3,,,,
qk+1
Q1=3,
Q2=7, Q3=31, Q4=211, Q5=2311, Q6=59*509, Q7=19*97*277, Q8=347*27953,
Q9=317*7037
63,
Q10=331*571*34231
The
number five is a prime number, after the number five is a composite number,
there is now a question of whether there are infinitely many k such that Qk is
a prime number, or the Qk number?
Look
at a matrix:
In
1934, East India (now Bangladesh) Chandra proposed a square sieve:
The
first line is the first of 4, 3 of the tolerance of arithmetic sequence 4, 7,
10,... , 4+3 (n1),...
The
second line is the first of 7, 5 of the tolerance of arithmetic sequence 7, 12,
17,... , 7+5 (n1),...
The
third line is the first of 10, 7 of the tolerance of arithmetic sequence 10,
17, 24,... , 10+7 (n1),...
The
fourth line is the first of 13, 9 of the tolerance of arithmetic sequence 13,
22, 31,... , 13+9 (n1),...
...
Line
m is the first 3m+1, tolerance of arithmetic sequence 2m+1 3m+1, 5m+2, 7m+3,...
, its n item is 3m+1+ (n1) (2m+1) =2mn+m+n,...
Write
the following array:
First
column, second column, third column, fourth column, fifth column... Column n...
First
row 47101316... 4+3 (n1)...
Second
row 712172227... 7+5 (n1)...
Third
row 1017243138... 10+7 (n1)...
Fourth
row 1322314049... 13+9 (n1)...
Fifth
row 1627384960... 16+11 (n1)...
...
... ... ... ... ... ... ... ...
Line
3m+1 5m+2 7m+3 9m+4 11m+5 m... 2mn+m+n...
...
... ... ... ... ... ... ... ...
The
secret lies in this square: if a natural number N appears in the table, then 2N
+ 1 is certainly not a prime number; if N does not appear in the table, then 2N
+ 1 is certainly a prime number.
In
fact, if N=2mn+m+n, then 2N+1=2 (2mn+m+n) +1=4mn+2m+2n+1= (2m+1) (2n+1), it is
not prime. On the contrary, the N does not appear in the table, if 2N+1 is not
a prime number, then 2N+1 must be a product of two odd, write 2N+1= (2m+1)
(2n+1) =4mn+2m+2n+1, N=2mn+m+n, it appears in the table, with the assumption of
contradiction. So when N does not appear in the matrix when 2N+1 is a prime
number.
Qk=q1*q2*q3,,,,
*qk+1, q1=2, so Q2=2* (q2*q3,,, QK) +1, apparently q2*q3*q4,,,, *qk
Is
the number, so q2*q3*q4, *qk=2N+1, N, and the above parties appear in the
screen, appearing in square sieve in N has an infinite number if the 2N+1
appear or do not appear in the matrix. Then it has an infinite number of the
Qk=2* (2N+1) +1 is prime or composite.
So
this conjecture is likely to be established!
Cite this
Article: Zhou M;
Hong WS (2019). A number of conjectures are likely to setup. Greener
Journal of Science, Engineering and Technological Research, 9(1): 67, http://doi.org/10.15580/GJSETR.2019.1.040919067. 