Given two integers n (n>1) and an odd prime p. Without loss of generality let p=2k1.
Prove that if C(n,2)  C(k,2) is divisible by p, it must be divisible also by p^{2}.
k=(p+1)/2
let F(n,k)=C(n,2)C(k,2)=n(n1)/2k(k1)/2=n(n1)/2(p+1)(p1)/4
so now let G(n,p)=(2n^22np^2+1)/4
if G(n,p)=0 mod p then we have
(2n^22np^2+1)/4=0 mod p
since p is odd, 4 has a multiplicitve inverse and thus can be canceled out
2n^22np^2+1=0 mod p (1)
2n^22n+1=0 mod p
now let G(n,p)=r mod p^2
and let 4 have a multiplicative inverse of i
then we get
2n^22np^2+1=ir mod p^2
2n^22n+1=ir mod p^2
from (1) above we now get
0=ir mod p^2
now since the multiplicative inverse of 4 can't be 0 we are left with
r=0 mod p^2
thus G(n,p)=0 mod p^2
thus if C(n,2)C(k,2) is divisible by p, then it is also divisible by p^2

Posted by Daniel
on 20170109 07:42:29 