Brocard’s Conjecture Mantzakouras Nikos, May 2015 Brocard conjecture in 1904 that the only solution of 1 ! 2 m n are n=4,5,7. There are no other solutions with 9 10 n .(Berndt and Galway n.d).Another of Brocard’s conjecture is that there are at least four primes between the squares of any two consecutive primes ,with the exception of 2 and 3.This related to Schinzel’s conjecture that,provided x is greater than 8,there is a prime between x and 2 ) (log x x .(See Opperman’s conjecture),[1]. The diophantine equation 1 2 m n! Initial solutions for n<=5 I) First we need to calculate two initial solutions when n = 4 and n = 5. Like Applies easily calculated , if we put m=2x+1 Z x into in Brocard equation and then we have the relation ) 1 ( 4 1 ! 2 x x m n (1) . This because for these values of n we have m odd number. Similarly we find ) 1 ( 4 1 ' )! 1 ( 2 y y m n (2), by Z y .From relations (1),(2) we conclude that ) 1 ( ) 1 ( ) 1 ( y y x x n (3). II) From(2&3) if call y=(n+1)=> 2) (n 4 n! ) 2 )( 1 ( 4 )! 1 ( n n n .The last has solution n=4 and m=5. III) Also from (2) we have if 1 / n n and / n y then 5 n 1) (n n 4 ! n / / / / and m=11. Therefore prove for values where n=4 and n=5. Generalization for n>5 From the original equation 1) (m 1) - (m n! 1 2 m n! then if m=2k+1, Z k i.e 1) (k k 4 1) (m 1) - (m n! .Also if from q p 5 4 3 2 n! with Z q p, and 1) (k k 4 q) (5 p) (6 4 q p 5 4 3 2 n! .(4), that means the system.. Z q p p k q k , , 6 1 5 From the previous system resulting equation 1 q 5 - p 6 .But should apply if we have w p then 1 w q and 6 w 1 1) (w 5 - w 6 and 7 1 w . More specifically examine two cases : If 1 = q 5 - 6 p has solutions k 5 + 1 = p and k 6 + 1 = q , Z k and if 1 - = q 5 - p 6 has solutions k 5 + 4 = p and k 6 + 5 = q . The first solutions it is true, because if k=1 ,p=6 && q=7 .
Brocard conjecture in 1904 that the only solution of n!=m^2-1 are n=4,5,7. There are no other solutions with n
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BrocardsConjecture Mantzakouras Nikos, May 2015 Brocard
conjecture in 1904that the only solution of1 !2 = m nare n=4,5,7.
There are no other solutions with910 < n.(Berndt and Galway
n.d).Another of Brocards conjecture is that there are at leastfour
primes between the squares of any two consecutive primes ,with the
exception of 2 and 3.This related to Schinzels conjecture
that,provided x is greater than 8,there is a prime between x and 2)
(log x x + .(See Oppermans conjecture),[1]. The
diophantineequation12m n! = Initial solutions for n
