Prime between Squares
There is at least one prime number between every pair of successive squares.
Proof:
Suppose n**2 -> (n+1)**2 all composite.
Then:
p0 | n**2 [p0 <= n]
p1 | n**2+1 [p1 <= n]
p2 | n**2+2 [p2 <= n]
…
p(2n) | n**2+2n [p(2n) <= n]
Note that:
hcf (pm, p(m+M)) = hcf (pm, M)
=> hcf (pm, p(m+P)) = hcf (pm, P) = 1 or P
Therefore, either pm, p(m+P) both divisible by P => both equal to P,
or, pm <> p(m+P)
Let: pm = 2 (or similar)
then: p(m+2) = 2
p(m+3) = 3/5/7/11…
p(m+5) = 5/7/11/13…
p(m+7) = 7/11/13/17…
p(m+11) = 11/13/17/19…
….
p(m+n)
using all primes <= n
=> p(m+1) > n
=> contradiction, from which result follows.