August 9th, 2018, 02:15 PM  #1 
Newbie Joined: Mar 2017 From: Norway Posts: 8 Thanks: 0  Please help
I am struggling to solve these simultaneous equations: $a^2+ab+b^2=7$ $b^2+bc+c^2=28$ $c^2+ac+a^2=21$ Any help would be highly appreciated. 
August 9th, 2018, 03:51 PM  #2 
Newbie Joined: Mar 2017 From: Norway Posts: 8 Thanks: 0 
I couldn't go any further than this: $(a+b)^2=7+ab$ $(b+c)^2=28+bc$ $(a+c)^2=21+ac$ 
August 9th, 2018, 04:14 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 413 Thanks: 227 Math Focus: Dynamical systems, analytic function theory, numerics 
Sum the first and third equation. Notice the LHS of each must sum to the LHS of the 2nd equation.

August 9th, 2018, 04:24 PM  #4 
Newbie Joined: Mar 2017 From: Norway Posts: 8 Thanks: 0 
$a^2+ab+b^2+c^2+ac+a^2=28$ $a^2+ab+b^2+c^2+ac+a^2=b^2+bc+c^2$ $2a^2+ab+acbc=0$ and now what 
August 9th, 2018, 04:34 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,852 Thanks: 749 Math Focus: Wibbly wobbly timeywimey stuff.  
August 9th, 2018, 04:39 PM  #6 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 
This problem has basically been answered over at FMH. https://www.freemathhelp.com/forum/t...73Pleasehelp It can be turned into a quartic. if the assumption is further made that the answers are integers, it can be shown that $b < 4.$ I gave one real answer over there, so there is at least one other real answer. Last edited by JeffM1; August 9th, 2018 at 04:41 PM. 
August 10th, 2018, 01:20 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,124 Thanks: 714 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
a=1, b=2, c=4 I got this answer by making the assumption that the numbers were all positive and then guessing some numbers to feel out the problem though, not by implementing a clever method or anything. Last edited by Benit13; August 10th, 2018 at 02:09 AM. 
August 10th, 2018, 05:36 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,285 Thanks: 1681 
7(a  b) + 28(b  c) + 21(c  a) = a³  b³ + b³  c³ + c³  a³ = 0 Hence c = 3b  2a. Substituting for c in the 3rd equation gives 9b²  12ab + 4a² + 3ab  2a² + a² = 21, so 3b²  3ab + a² = 7. Subtracting that from the first equation gives 4ab  2b² = 0. Hence 2b(2a  b) = 0. Substituting b = 2a into the first equation gives a² = 1. Hence (a, b, c) = (1, 2, 4) or (1, 2, 4) or (√7, 0, 2√7) or (√7, 0, 2√7). 
August 10th, 2018, 08:25 AM  #9  
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  Quote:
 
August 10th, 2018, 09:54 AM  #10  
Senior Member Joined: Sep 2016 From: USA Posts: 413 Thanks: 227 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
In reality, despite the large amount of structure in this problem, it is still extremely difficult or impossible to solve most nonlinear systems exactly. Even in the case you know ahead of time that there are integer solutions, trying to eliminate variables is still the worst possible way to proceed. In this case you would just do Newton and round to the nearest integer. There is no case where playing around with the equations is the "right" thing to do yet at low level this is the only thing students are equipped to do. These problems are toxic.  