 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Floor and Square Division Settlement (Posted on 2016-01-01) 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) Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information