grimm'sconjecture

Grimm's Conjecture

For any run of m consecutive composite numbers, there exist distinct prime divisors of each.

Proof:

1) observe that m^2<2n, By Prime between Squares [start of run at n+1]

2) hcf (pj, p(j+k)) = hcf (pj, k)

therefore either pj, p(j+k) both | k or pj <> p(j+k)

3) now introduce qj etc. also:

either qj, q(j+k) both | k or qj <> q(j+k)

4) but k<= m =>

p(j+k), q(j+k) can't both simultaneously | k, since one must be > k, hence, WLOG

5) pj <> p(j+k), for all k