please reply for the proof given

[ram27]

sir/madam this is my finding on J.P Jones conjecture...the conjecture
states
as follows...

if (2n-1)C(n-1) congruent to 1 mod n^3 then n is prime.....

my solution towards it......

consider,
2n-1Cn-1 = (2n-1)!/n! * (n-1)!
= {(2n-1)(2n-2)(2n-3)....n(n-1)(n-2)
(n-3)...3.2.1} /{ n! * (n-1)!}
canceling out n! we get

= {(2n-1)(2n-2)(2n-3)....(n+1)} /
{(n-1)!}

simplifying the above expression we get,

= 2^((n-1)/2)*{(2n-1)(2n-3)(2n-5)....}/
{(n-1/2)!}

now dividing this by n^3 (i.e mod n^3) we see that the above
expression satisfies only for prime values.....

the expression 2^(n-1/2) divided by n will result in 1 or -1 only for
prime values of n...the remaining is taken care of the
expression....so this proves it....i.e

= [{2^(n-1/2)}/n ]*P/Q

where P is the product of all odd values while Q is the product of n^2
and (n-1/2)!.....

the expression 2^((n-1)/2) is a consequence of Euler's psuedo
prime..replacing a by 2..i.e

aCp congruent to a^(p-1)/2 mod p....

please do check it and verify it for me......

Regards

Ram