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.)
Some Thoughts Partial solution | Comment 2 of 4 |
From the equation, we know that sē must be odd, since the term (4p4+ 4p3+ 4p2 + 4p) is even.  Therefore, s must also be odd.

We also know that for every odd s:       (s+1)(s-1) is evenly divisible by 4.

The obvious solutions are:

{(0,-1), (0,1), (2,-11), (2,11)}

However, as p increases, something ppeculiar happens.

(4p4+ 4p3+ 4p2 + 4p) = i + r

where i = an integer & r = the remainder.  The solution of course is to find a p where r = 0.  However, as p→∞, it appears that r→0.75.  If this holds true, then there are no more solutions for this equation.

  Posted by hoodat on 2007-05-14 13:25:19
Please log in:
Remember me:
Sign up! | Forgot password

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

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information