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

Home > Just Math
Symmetry helps! (Posted on 2016-10-26) Difficulty: 4 of 5
Solve the system of equations below, listing all un-ordered triplets of integers:

x^2 −(y+z+yz)x+ (y+z)yz = 0
y^2 −(z+x+zx)y+ (z+x)zx = 0
z^2 −(x+y+xy)z+ (x+y)xy = 0

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re: my short solution | Comment 6 of 7 |
(In reply to my short solution by Ady TZIDON)

there are 3 missing solutions, namely when two values are -1 and the third is 1


(-1,-1,1), (-1,1,-1), and (1,-1,-1)

this corresponds to x=yz, y=xz, and z=xy
using the first two we get
x=xz^2
if x=0 then we have y=0 and z=0 which we already have
if x!=0 then we have
z^2=1

case 1:
z=1
then x=y and xy=1 giving x^2=1 thus x=1 or x=-1
x=1 we already have but x=-1 gives us (-1,-1,1)

case 2:
z=-1
then x=-y and xy=-1 giving -x^2=-1 x^2=1 thus x=-1 or x=1
this gives us the other 2 solutions (-1,1,-1) and (1,-1,-1) 

  Posted by Daniel on 2016-10-27 05:29:47
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 (6)
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