Find all possible positive integers P, which are not perfect squares, such that (floor(√P))3 divides P2
let n = floor(sqrt(P))
Then the problem becomes one of finding all integer (n,P) such that
n^2 < P < (n+1)^2
and n^3 divides P^2
But n^3 divides P^2 only if n divides P,
which means that P is either n(n+1) or n(n+2)
(because P is between n^2 and (n^2+2n+1), non-inclusive).
If P = n(n+1), then P^2/n^3 = (n+1)^2/n = n + 2 + 1/n, so n can only be 1.
If P = n(n+2), then P^2/n^3 = (n+2)^2/n = n + 4 + 4/n, so n can only be 1,2 or 4
So the only P that work are
P = 1*(1+1) = 2
P = 1*(1+2) = 3
P = 2*(2+2) = 8
P = 4*(4+2) = 24
Analytic solution (spoiler)
