Brocard Conjecture

1. Brocard’s Conjecture
Mantzakouras Nikos, May 2015
Brocard conjecture in 1904 that the only solution of 1! 2
 mn 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 xx  .(See Opperman’s conjecture),[1].
The diophantine equation 12mn! 
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 
Zx into in Brocard
equation and then we have the relation )1(41! 2
 xxmn (1) . This
because for these values of n we have m odd number. Similarly we
find )1(41')!1( 2
 yymn (2), by 
Zy .From relations (1),(2) we
conclude that )1()1()1(  yyxxn (3).
II) From(2&3) if call y=(n+1)=> ))2)(1(4)!1( 2(n4n!  nnn .The
last has solution n=4 and m=5.
III)Also from (2) we have if 1 nne
and e
ny  then
5n1)(nn4 eee
)!(ne
and m=11. Therefore prove for values
where n=4 and n=5.
Generalization for n>5
From the original equation 1)(m1)-(mn!12mn!  then if m=2k+1,

Zk i.e 1)(kk41)(m1)-(mn!  .Also if from qp5432n!  with

Zqp, and 1)(kk4q)(5p)(64qp5432n!  .(4), that means the
system..









Zqp
pk
qk
,,
61
5
From the previous system resulting equation 1q5-p6  .But should apply if
we have wp  then 1wq  and 6w11)(w5-w6  and 71w  .

2. More specifically examine two cases: If 1=q5-6  p has solutions k5+1=p and
k6+1=q , 
Zk and if 1-=q5-p6  has solutions k5+4=p and k6+5=q . The
first solutions it is true, because if k=1 ,p=6 && q=7 .
In second case p=4 && q=5 we already know and is truth for k=0, but not for our
case ,because we want n> 5, therefore only exist the case 1=q5-p6  . Also we
need 
Zk and 1qp then we have if 1=q5-p6  => kqp  (5) and too
if 1-=q5-p6  then 1 kqp (6).The satisfaction of two cases occurs for the
first k= 1, and also the second by k = 0. But the first is the desired and
acceptable.
The case m=2k, 
Zk does not exist ,because the right part of the eq. 12mn! 
is odd and the left even which is impossible thing. To this end therefore accept
for n>5 only n = 7 and m=71,without another n order to comply, with the
criterion of factorial(n!), that the next number to be increased to the previous
1 unit. We see therefore that the Brocard conjecture has only solution of
1! 2
 mn the values n=4,5,7.
Bibliography
[1].The most Mysterious figures in Math. David Wells.
[2].Brocard’s problem and variations Yi Liu.