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 910n .(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 eny 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 .

More specifically examine two cases: If 1 = q5 - 6 p has solutions k 5 + 1 = p and

k 6 + 1 =q , Zk and if 1- = q5 - p6 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 . 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 1qp 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.