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

Home > Just Math
Product + Square = Difference of Squares (Posted on 2011-07-03) Difficulty: 3 of 5
Three positive integers P, Q and R, with P < Q < R, are in arithmetic sequence satisfying :
N*P*Q*R + Q2 = R2 - P2, 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.

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.)
Possible solution | Comment 1 of 2

1.  N*P*Q*R + Q^2 = R^2 - P^2:

I   Let Q=p;Let P=(p-k); Let R=(p+k)
II  n*p*(p-k)*(p+k)+p^2=(p+k)^2-(p-k)^2
III np^3+p^2-k^2np = 4kp Expanding, simplifying
IV  p(np+1) = k(kn+4)        Cancelling p throughout
When  {k,n,p} = {2,1,3}

2.  Solving for successive n reveals that {k}={-6/n, 2/n}, {p}={-4/n,3/n}, so {k,n,p} = {2,1,3} is unique in the positive integers.


  Posted by broll on 2011-07-04 02:08:01
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 (3)
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