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Floor and Square Division Settlement (Posted on 2016-01-01) Difficulty: 3 of 5
Find all possible positive integers P, which are not perfect squares, such that (floor(√P))3 divides P2

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Analytic solution (spoiler) Comment 4 of 4 |
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

  Posted by Steve Herman on 2016-01-03 08:06:42
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