Problems & Puzzles:
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 =
(Ref. 2, p. 43)
Dagfinn Nordås, from Norway
(13/02/99) has an explanation to the last paragraph of
this Problem. Here is his e-mail:
"In problem 2 you ask if Pomerance`s observation
has anything to do with the "peculiar" fact
My answer: 210-p (210-p<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 (210-107), 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,210-p (n/2<p<n) is prime." E-mail: email@example.com
Said El Aidi ,from Morocco wrote (13/06/2000):
My answer is:
"In problem 1 you ask if we can construct a similar
interesting question like or related with this of Pomerance".
N=prime=n-p for all 13n/14<p=prime<n-1 ,only for n=2310
I verified that 2310-p 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<n-1.
This has nothing to do with the fact that n=2*3*5*7*11. My e.mail:
Two days later, Said El Aidi wrote again the
"I found a generalization of Carl Pomerance Observation,
to the two questions at the same time.
This is my result :
For every integer n=>210.
N=prime =n-p for all n*[n/p²k+1]/([n/p²k+1]+1)<p =prime<n-1,
only for n =pk# =2*3*5*7*…*pk.
(Here [x] is the greatest integer <= x , usually called the "floor function").
Let p be a prime such that n*[n/p²k+1]/([n/p²k+1]+1)<p<n-1
We have N=n-p< n/([n/p²k+1]+1) <p²k+1
Suppose N is composite.Then N must have a prime factor lesser
than pk+1 (2,3,5,7,…,or pk).
As N and p are relatively prime (Because N=n-p<p),therefore
p=n -N is composite. This contradicts the fact that p is prime.
Then N must be prime.
1. For n =p4 # =210 , 210-p is prime for all the primes p
such that n/2<p<n-1.
2. For n =p5 # =2310 , 2310-p is prime for all the primes p
such that 13n/14<p<n-1.
3. For n =p6 # =30030 , 30030-p is prime for all the primes p
such that 103n/104<p<n-1."