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August 9th, 2018, 02:15 PM   #1
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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.
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August 9th, 2018, 03:51 PM   #2
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I couldn't go any further than this:
$(a+b)^2=7+ab$
$(b+c)^2=28+bc$
$(a+c)^2=21+ac$
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August 9th, 2018, 04:14 PM   #3
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Sum the first and third equation. Notice the LHS of each must sum to the LHS of the 2nd equation.
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August 9th, 2018, 04:24 PM   #4
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$a^2+ab+b^2+c^2+ac+a^2=28$
$a^2+ab+b^2+c^2+ac+a^2=b^2+bc+c^2$
$2a^2+ab+ac-bc=0$
and now what
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August 9th, 2018, 04:34 PM   #5
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Quote:
Originally Posted by Farzin View Post
$a^2+ab+b^2+c^2+ac+a^2=28$
$a^2+ab+b^2+c^2+ac+a^2=b^2+bc+c^2$
$2a^2+ab+ac-bc=0$
and now what
This may or may not be what SDK is moving you toward but notice that you can solve for c in terms of a and b in your last equation.

-Dan
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August 9th, 2018, 04:39 PM   #6
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This problem has basically been answered over at FMH.

https://www.freemathhelp.com/forum/t...73-Please-help

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.
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August 10th, 2018, 01:20 AM   #7
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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.
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August 10th, 2018, 05:36 AM   #8
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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).
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August 10th, 2018, 08:25 AM   #9
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Quote:
Originally Posted by skipjack View Post
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).
Very pretty. Once you "see" the difference of cubes, you can avoid the quartic altogether.
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August 10th, 2018, 09:54 AM   #10
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Quote:
Originally Posted by topsquark View Post
This may or may not be what SDK is moving you toward but notice that you can solve for c in terms of a and b in your last equation.

-Dan
Nope I was in a hurry and made a mistake. I thought the sum had a factor of $a$. I strongly dislike questions like this at a lower level since it gives the impression that math is just a bunch of clever tricks to be remembered.

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.
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