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

Home > Just Math
Product + Square = Difference of Squares (2) (Posted on 2011-07-20) Difficulty: 3 of 5
Following on from this problem, three positive integers P, Q and R, (from smallest to largest in that order), are in arithmetic sequence satisfying : N*P*Q*R + Q = R^2 - P^2, where N is a positive integer.

Determine all possible quadruplet(s) (P, Q, R, N) that satisfy the above equation, and prove that no other quadruplet satisfies the given conditions.

Note that in this variant, the second term involving Q is not a square.

See The Solution Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2
I believe the product NP is constrained much as in the original problem, so arguing along the same lines...

NPQR + Q = R2 - P2  gives           NP = (R2 - P2 - Q)/(QR)

Writing  R = 2Q - P  and simplifying then gives      NP = (4Q - 4P - 1)/(2Q - P)

which can be rearranged to                                 NP = 2  -  (1 + 2P)/(2Q - P)

Since 2Q > P, it follows that NP < 2 and, since N and P are both positive integers, we have N = P = 1 as the only possibility.

Substituting then gives P, Q, R, N  to be 1, 2, 3, 1 respectively.



  Posted by Harry on 2011-07-25 19:11:41
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 (9)
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