1
Name:
Anonymous
2007-11-25 10:43
Proof that for every square > 1, there exists another square so that their difference is a semiprime.
2
Name:
Anonymous
2007-11-25 14:30
just accept it, somebody else probably already proved it. No sense in reinventing the wheel.
3
Name:
Anonymous
2007-11-25 15:01
>>2
It's probably his homework, dimwit.
5
Name:
Anonymous
2007-12-01 13:36
Guys I need this before monday :(
6
Name:
Anonymous
2007-12-01 14:19
Given a^2, we want b^2 so that (a^2 - b^2) = pq, so a-b = p, a+b = q. a - p = b = q - a, so 2a = p + q. Now just prove Goldbach's conjecture and you're finished.
7
Name:
Anonymous 2007-12-03 1:32
>>3
you are a delicate little flower, aren't you