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

Home > Just Math
Some Powers Sum Square II (Posted on 2013-12-18) Difficulty: 3 of 5
Determine all possible pairs (P, Q) of integers such that :
(19P + Q)18 + (P + Q)18 + (P + 19Q)18 is a perfect square.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
solution and happy new year | Comment 1 of 2
p=0, q=0 is a solution.

For p<>0, q=0, p^18 * (19^18 + 2) must be a square which implies (19^18 + 2) = (4k + 3) is a square which is impossible.

If the greatest common factor of (p,q) = m>1, we can factor m^18 from both sides of the equation to get an equation of the same form as the original.

So take m=1.  Then one of p,q is odd and the other even, making each term of the LHS odd and their sum = (4k + 3), again impossible.  

  Posted by xdog on 2014-01-01 01:25:20
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 (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information