Problems & Puzzles: Puzzles

 Puzzle 444. Pseudo twin primes Sebastián Martín Ruiz sent the following puzzle: Let n,m  positive integres greater than one. Let TW(n,m)=((n-1)!+1)((m-1)!+1)(n^2+m^2)/(n^2m^2+2nm) We define n and m are pseudotwinprimes iff  TW(n,m) is integer Questions:  1)   Prove that: If n,m are twin primes then n,m are pseudotwin primes. (easy) 2)   Two examples of  pseudotwinprimes (n, m) are: (7,191) and (41, 1993). Find more pairs of pseudotwinprimes  that don’t are twin primes

Contributions cam from Farideh Firoozbakht, David Broadhurst & Alexey Tetyorko,

Farideh wrote:

1. TW(n,n+2) = ((n-1)!+1)((n+1)!+1)(n^2+(n+2)^2)/(n^2(n+2)^2+2n(n+2))
= 2((n-1)!+1)((n+1)!+1)/(n(n+2))
= 2*(((n-1)!+1)/n)*(((n+1)!+1)/(n+2))

Now if both numbers n & n+2 are primes then by Wilson's Theorem
both numbers ((n-1)!+1)/n & ((n+1)!+1)/(n+2) are integers so if n &
n+2 are twin primes then n & n+2 are pseudotwin primes.

2. (59,13537) is a pair of pseudotwin primes that they aren't twin primes.

***

David wrote:

[13537, 59]
[45293, 241]

***

Alexey wrote:

Answer. Let n, m be primes, then we can rewrite TW(n,m) as follows
TW(n,m)=(((n-1)!+1)/n)*((m-1)!+1)/m)*((n^2+m^2)/(nm+2)),
now, the two leader factors are integers by the Wilson's theorem. Moreover, we can use m=n+2 (they are twin prime!), and
the last factor ((n^2+m^2)/(nm+2))=(n^2+(n+2)^2)/(n(n+2)+2)=(2n^2+4n+4)/(n^2+2n+2)=2. So TW(n,m) is an integer. QED.

***

 Records   |  Conjectures  |  Problems  |  Puzzles