b*k
发帖数: 27 | 1 Determine whether or not therer exists a positive integer n such that
n is divisible exactly 2000 different prime numbers, and
2^n+1 is divisible by n | l**i
发帖数: 5 | 2 I think k could be 3^n or 2*3^n.
It's easy to check that 9*19|2^{9}+1
Use induction to prove that p1^k*p2*...*pk|2^{3^k}+1 where p1=3,p2=19
with all the pi's distinct
we already have the case k=2,assume k>2 is correct,prove case k+1
because 2^{3^{k+1}}+1=(2^{3^k}+1){(2^{3^k})^2-2^{3^k}+1}
let A=(2^{3^k})^2-2^{3^k}+1,we have 3|A,A=3(mod pi)(i>1),so pi!|A(i>1)
assume 2^{3^k}=3m-1,then A=9m^2-9m+3=3*{3m*m-3m+1),let B=3m*m-3m+1
we see pi!|B for all i,so B must have a different prime factor p{k+1}
an |
|
