Problems & Puzzles: Puzzles

Puzzle 1176  ((((n-1)!)^2+4*n-1))/(n*(2n-1)) Davide Rotondo sent the following puzzle: If n*(2n-1) divides ((((n-1)!)^2+4*n-1)) then n is a prime p such that 2p-1 is prime ex. n=7 (((6!)^2+4*7-1))/(7*(7*2-1)) = 5697 Q. true or false?

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From May 16 to 24, contributions came from Flavio Torasso, Ken Wilke, Emmanuel Vantieghem

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Flavio Torasso wrote:

Puzzle 1176 is true according to Corollary 1 of the paper "Simultaneous primality of the integers n and 2n-d" available at
https://studmath-old.up.krakow.pl/index.php/studmath/article/view/148/114

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Ken Wilke wrote:

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Emmanuel wrote:

Put  F[n] = ((n - 1)!/^2 + 4 n - 1.

 

Davide's proposition sounds :

    n(2n -1)  divides  F(n)  ==> n  and  2n - 1  are prime.
 

If  n  divides  F(n)  and is composite, then it has a prime divisor  p  less than  Sqrt(n).  That  p  divides  ((n - 1)!  and  clearly  n, but not  F(n).  Contradiction !
Thus, if  n  divides  F(n), it cannot be a composite number (but, it can be  1).
If  2n - 1  divides  F(n)  and has a prime divisor  q  less than  Sqrt(2n - 1), then that  q  also divides  ((n - 1)!  but not  4n - 1.  Contradiction !
Thus, 2n - 1  divides  F(n) , it cannot be composite (but it can be  1).
Thus, Davide's proposition is true with the extra condition  n > 1..

Though it is not asked, it is of interest to examine the converse.
Suppose  n  and  2n - 1  are both prime.
By Wilson's theorem, n  divides  ((n - 1)!)^2 - 1  and  n.  Thus, it divides  F(n).
Applying Wilson's theorem for the prime  q = 2n - 1  we have  mod q :
   (q - 2)! = 1 = 1*2* ... *(n-1)*[n*(n+1* ...  *(q - 2)]
                     = (n - 1)!*[(-(n-1))(-(n-2))* ... *(-2)] = ((-1)^(n-2))((n-1)!)^2
Thus, when  n - 2  is odd, we have :
   ((n-1)!)^2 === -1 mod q.
Whence  F(n) === -1 + 4n - 1 = 2q ===0 mod q.

The converse is true.
When  n = 2 (i. e. : q = 3), we have trivially  F(2) = 8.

In that case the converse does not hold. 

 

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