Given two integers n (n>1) and an odd prime p. Without loss of generality let p=2k-1.
Prove that if C(n,2) - C(k,2) is divisible by p, it must be divisible also by p2.
k=(p+1)/2
let F(n,k)=C(n,2)-C(k,2)=n(n-1)/2-k(k-1)/2=n(n-1)/2-(p+1)(p-1)/4
so now let G(n,p)=(2n^2-2n-p^2+1)/4
if G(n,p)=0 mod p then we have
(2n^2-2n-p^2+1)/4=0 mod p
since p is odd, 4 has a multiplicitve inverse and thus can be canceled out
2n^2-2n-p^2+1=0 mod p (1)
2n^2-2n+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^2-2n-p^2+1=ir mod p^2
2n^2-2n+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
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Posted by Daniel
on 2017-01-09 07:42:29 |
proof
