Problems & Puzzles: Puzzles

Puzzle 1100 Another function to produce only and all the primes?

Sebastián Martín Ruiz sent the following prime-function:

Given the function:

P(n, k) = GCD[2n + 1, k! n! + 1] + Floor[1/GCD[2n + 1, k! n! + 1]]

where n>=1 and k takes only the following three values k=0,n,2n. Each prime greater than two appears only once, and the prime 2 repeats infinitely.

Is this a "serious" prime function producer?


Q1) The function P(n,k) ONLY generates primes.

Q2) The function P(n,k) generates ALL prime numbers or find a counterexample.

Q3) For a fixed n, if P(n,k) is prime greater than 2 only appears once for any k=0,n,2n

During the week 20-26 August 29022, contributions came from Giorgos Kalogeropoulos, Emmanuel Vantieghem


Giorgos wrote:

We are interested in the function   GCD[2n + 1, k! n! + 1] 
If 2n+1 is prime then it divides k!n!+1 for one of the three cases for k={0,n,2n} as we will explain below in the proof.
Because 2n+1 must be prime and it is smaller than k!n!+1 the gcd can only be 1 or 2n+1.

If 2n+1 is not prime then obviously any divisor of 2n+1 is less than n and cannot divide k!n!+1 for k={0,n,2n}, this means that gcd is 1. 
So, the results can only be 1 or p=2n+1 with p prime

The floor function is trivial and it just adds 1 if the result is 1 or 0 if the result is p.
It was added on purpose in order that all 1's become 2 and all primes stay the same.
We will examine the three different cases for k
a)k=0, gcd(2n+1,n!+1)
b)k=n, gcd(2n+1,(n!)^2+1)
c)k=2n, gcd(2n+1, 2n!*n!+1)

If 2n+1=p then n=(p-1)/2 and we can rewrite the above three cases as follows:
a)k=0, gcd(p, ((p-1)/2)!+1)
b)k=n, gcd(p,  (((p-1)/2)!)^2+1)
c)k=2n, gcd(p, (p-1)! ((p-1)/2)! +1)

we will examine the conditions needed for p to divide the second part of the gcd in the above cases.
In the book ``History of the Theory of Numbers,'' Vol. 1 (p. 275)(Dickson 1919) we can find the following
We can derive from Wilson's theorem that for prime p
(1) (((p-1)/2)!)^2 = (-1)^((p+1))/2) mod p
If we use this in case (b) for k=n we see that
p divides (((p-1)/2)!)^2+1 only if (p+1)/2 is odd

 (p+1)/2 = 2m+1 which means p=4m+1

This means that case (b) covers all primes congruent to 1 mod 4.
If the other two cases (a) and (c) cover the primes congruent to 3 mod 4 in a unique way, then we are finished.
In the same page of that book, it is written:
for p prime of the form 4n+3, we have
((p-1)/2)!=±1 mod p
So, the primes of the form 4n+3 can be divided into two subgroups according to ((p-1)/2)! being congruent to 1 or -1 mod p.
In case (a) with k=0 we can easily see that there belong the primes p such that ((p-1)/2)! = -1 mod p.
The only thing that is left to prove is that the primes of the form 4n+3 with ((p-1)/2)! = 1 mod p belong in case (c).
((p-1)/2)! = 1 mod p (1A)
From Wilson's theorem we have
(p-1)!= -1 mod p for any prime p (1B)
if we multiply (1A) and (1B) we get
 (p-1)! ((p-1)/2)! = -1 mod p and p divides  (p-1)! ((p-1)/2)!+1 which is exactly our case (c).

So, gcd(p, (p-1)! ((p-1)/2)! +1) = p iff p is prime of the form 4n+3.
In conclusion, this function returns {1,1,1} if 2n+1 is not prime, and divides the primes in three groups if 2n+1 is prime.
Primes of the form 4n+1 are covered by the second case (b) for k=n.
Primes of the form 4n+3 are covered by cases (a) and (b).
This function generates all primes with no duplicates.

But it is trivial as a result of Wilson's theorem, so it is not a "serious" prime generating function


Emmanuel wrote:

If  2n + 1  is not prime, it is divisible by a prime less than  n, whence  GCD[2n + 1, k!*n! + 1] = 1  and hence  F(n,k) = 2  for k = 0, n, 2n.
If  2n + 1 = p  is prime, then; by Wilson's theorem, it divides  (2n)! + 1.
But we have : (2n)! + 1 = 2n*(2n-1)*...*(n+1)*n! + 1 === ((-1)^n)*(n!)(n!) + 1 (mod  p).
 * If  n  is odd, (1 + n!)(1 - n!)  is divisible by  p.  Thus :
      either  (1 + n!)  is divisible by  p  which results in  F(n,0) = p.
         We then have  F(n, n) = 2  and  F(n, 2n) = 2
      or : n! - 1  is divisible by  p.  In that case, F(n,0) = 2, F(n, n) = 2  and  F(n, 2n) = p
 * If  n  is even then we have   F(n, n) = p  and  F(n,0) = 2 = F(n,2n).

No doubt, this is a "serious" prime producer.
If  2n + 1  is not prime, it is divisible by a prime less than  n, whence  GCD[2n + 1, k!*n! + 1] = 1  and hence  F(n,k) = 2  for k = 0, n, 2n.

Of course, the computation of  n!  becomes very slow with growing  n.
For instance, Computing  G(10^9)  gives  2  after more than half an hour.  In much less time we can detect that  2*10^9 + 1 is divisible by  3 ...


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