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

Home > Just Math
Quadratic and Sextic (Posted on 2015-06-04) Difficulty: 3 of 5
Find all possible triplets (A,B,C) of positive integers that satisfy the following system of equations:

A2 = 2(B+C) and:
A6 = B6 + C6 + 31(B2 +C2)

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Answer Comment 1 of 1
If you cube the first equation you can substitute Ain the second:
8(B+C)3 = B6 + C6 + 31(B2 +C2)

When I entered this into Wolfram|Alpha the graph is a rather small egg shape with the blunt end at b=1, c=1 and narrow end around b= 2.75, c=2.75

There are solutions very near (2,1) and (1,2) but not quite.

The only positive integer solution is (A,B,C)=(2,1,1).

  Posted by Jer on 2015-06-05 11:08:59
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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information