primebetweensquares

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.