All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Square An Integer, Get Quartic (Posted on 2007-05-13) Difficulty: 3 of 5
Analytically determine all possible integer pairs (p, s) such that:
4p4+ 4p3+ 4p2 + 4p +1 = s2.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Squeeze Play (Solution) | Comment 3 of 4 |

If p satisfies the compound inequality (2p^2 + p + 1)^2 > 4p^4 + 4p^3 + 4p^2 + 4p + 1 > (2p^2 + p)^2 then it is obvious that the middle expression is not a square since it is strictly between two consecutive squares.

The only intervals where the compound inequality fails are [-1, -1/3] and [0, 2].  There are only four integers in this range: -1, 0, 1, and 2.

Direct calculation shows that p=1 is not a solution and p=-1 and p=0 are solutions with s = +/-1 and p=2 is a solution with s=+/-11.

The complete solution set is (-1, 1), (-1, -1), (0, 1), (0, -1), (2, 11), (2, -11).


  Posted by Brian Smith on 2007-05-14 23:02:03
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 (16)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information