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