Problems & Puzzles:
Problems
Problem 2. Carl Pomerance Observation
N = prime = n  p for all n/2<p
= prime < n, only for n = 210
Pomerance observed that 210  p is
prime for p =107, 109, 113, 127, 131, 137, 139, 149, 151,
157, 163, 167, 173, 179, 181, 191, 193, 197 & 199,
being p all the primes such that n/2<p< n.
He asked for another number n
greater than 210 with the same property but "With
the help of Deshouillers, Granville & Narkiewicz, he
later answered this negatively".
1. can you construct a similar
interesting question like or related with this of Pomerance?.
2. Has anything to do with the
Pomerance observation, the peculiar fact that 210 =
2*3*5*7 ?
(Ref. 2, p. 43)
Solution
Question 2.
Dagfinn Nordås, from Norway
(13/02/99) has an explanation to the last paragraph of
this Problem. Here is his email:
"In problem 2 you ask if Pomerance`s observation
has anything to do with the "peculiar" fact
that 210=2*3*5*7?
My answer: 210p (210p<11^2, p is prime) will
always be prime because since 2,3,5,7 goes up in 210 it
can`t go up in 210 +/ p.
Since the biggest number here is 103 (210107), it must
have a prime factor lesser than 11, and since these
primes are 2,3,5,7 which I have proven(hopefully) is not
a factor,210p (n/2<p<n) is prime." Email: dagfinnn@hotmail.com
Question 1.
Said El Aidi ,from Morocco wrote (13/06/2000):
"In problem 1 you ask if we can construct a similar
interesting question like or related with this of Pomerance".
My answer is:
N=prime=np for all 13n/14<p=prime<n1 ,only for n=2310
I verified that 2310p is prime for p=2153,2161,2179, 2203, 2207
,2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281,2287,2293
& 2297, being p all the prime such that 13n/14<p<n1.
This has nothing to do with the fact that n=2*3*5*7*11. My e.mail:
Saidofski@hotmail.com "
***
Two days later, Said El Aidi wrote again the
following:
"I found a generalization of Carl Pomerance Observation,
answer
to the two questions at the same time.
This is my result :
For every integer n=>210.
N=prime =np for all n*[n/p²_{k+1}]/([n/p²_{k+1}]+1)<p =prime<n1,
only for n =p_{k}# =2*3*5*7*…*p_{k}.
(Here [x] is the greatest integer <= x , usually called the "floor function").
Proof :
Let p be a prime such that n*[n/p²_{k+1}]/([n/p²_{k+1}]+1)<p<n1
We have N=np< n/([n/p²_{k+1}]+1) <p²_{k+1}
Suppose N is composite.Then N must have a prime factor lesser
than p_{k+1} (2,3,5,7,…,or p_{k}).
As N and p are relatively prime (Because N=np<p),therefore
p=n N is composite. This contradicts the fact that p is prime.
Then N must be prime.
Example:
1. For n =p_{4} # =210 , 210p is prime for all the primes p
such that n/2<p<n1.
2. For n =p_{5} # =2310 , 2310p is prime for all the primes p
such that 13n/14<p<n1.
3. For n =p_{6} # =30030 , 30030p is prime for all the primes p
such that 103n/104<p<n1."
***
