MHB Solving for Real Solutions: Systems of Equations with Powers

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The discussion revolves around solving a system of equations involving variables a, b, c, and d. Two solutions are identified: (-2, -1, -1, 1) and (1, 2, -1, 1). Participants confirm the validity of these solutions through various algebraic manipulations and substitutions. The approach includes rewriting equations for clarity and deriving relationships between the variables. Ultimately, the consensus is that these are the only solutions to the problem.
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Solve for real solutions of

$a-b-c+d=1$

$a^2+b^2-c^2-d^2=3$

$a^3-b^3-c^3+d^3=-5$

$a^4+b^4-c^4-d^4=15$
 
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The only way I can do this is by guesswork.
[sp]$a-b-c+d=1 = (-2) +3$

$a^2+b^2-c^2-d^2=3 = (-2)^2 - 1$

$a^3-b^3-c^3+d^3=-5 = (-2)^3 + 3$

$a^4+b^4-c^4-d^4=15 = (-2)^4 - 1$.

When you write them like that it's easy to see that $(a,b,c,d) = (-2,-1,-1,1)$ is a solution. But is it unique?

Edit (in haste). It looks as though $(a,b,c,d) = (1,2,-1,1)$ is another solution.[/sp]
 
Last edited:
My solution
I'll solve the first equation for $c$ so

$c =a-b+d-1$

This give the second equation

$2\,ab-2\,ad+2\,a+2\,bd-2\,b-2\,{d}^{2}+2\,d-4=0$

or solving for $b$

$b = {\dfrac {ad-a+{d}^{2}-d+2}{a+d-1}}$.

Note that $a+d-1 \ne 0$ since if this was true, there's no solution to the system. With these two assignments, the remaining equations become

$3\,{\dfrac {{a}^{2}d+{a}^{2}+a{d}^{2}-ad+2\,a-4+2\,d-2\,{d}^{2}}{a+d-1}
}
=0\;\;\;\;\;(1)$
$4\,{\dfrac { \left( d+1 \right) \left( a-2 \right) \left( a+d
\right) \left( {a}^{2}+a-{d}^{2}+2\,d-3 \right) }{ \left( a+d-1
\right) ^{2}}}
=0\;\;\;\;\;(2)$

From $(2)$ we see 4 possibilities: $d=-1,a=2$ or $a+d=0$. However, each one of these gives that $(1)$ cannot be satisfied so we are left with

${a}^{2}+a-{d}^{2}+2\,d-3=0$

together with

${a}^{2}d+{a}^{2}+a{d}^{2}-ad+2\,a-4+2\,d-2\,{d}^{2}=0$

Eliminating $d^2$ gives

$(a+2)(a-1)(a+d-1)=0$

which leads us to $a = 1$ and $a = -2$ from which we can find the remaining variables $b, c$ and $d$ leading to the solutions that Opalg found.
 
Opalg said:
The only way I can do this is by guesswork.
[sp]$a-b-c+d=1 = (-2) +3$

$a^2+b^2-c^2-d^2=3 = (-2)^2 - 1$

$a^3-b^3-c^3+d^3=-5 = (-2)^3 + 3$

$a^4+b^4-c^4-d^4=15 = (-2)^4 - 1$.

When you write them like that it's easy to see that $(a,b,c,d) = (-2,-1,-1,1)$ is a solution. But is it unique?

Edit (in haste). It looks as though $(a,b,c,d) = (1,2,-1,1)$ is another solution.[/sp]

Thanks for participating, Opalg!:)

I like the way you have the given values of the 4 equations rewritten as $(-2) +3$, $(-2)^2 - 1$, $(-2)^3 + 3$, $(-2)^4 - 1$. And yes, these two are the only solutions to the problem. Good observation, Opalg!

Jester said:
My solution
I'll solve the first equation for $c$ so

$c =a-b+d-1$

This give the second equation

$2\,ab-2\,ad+2\,a+2\,bd-2\,b-2\,{d}^{2}+2\,d-4=0$

or solving for $b$

$b = {\dfrac {ad-a+{d}^{2}-d+2}{a+d-1}}$.

Note that $a+d-1 \ne 0$ since if this was true, there's no solution to the system. With these two assignments, the remaining equations become

$3\,{\dfrac {{a}^{2}d+{a}^{2}+a{d}^{2}-ad+2\,a-4+2\,d-2\,{d}^{2}}{a+d-1}
}
=0\;\;\;\;\;(1)$
$4\,{\dfrac { \left( d+1 \right) \left( a-2 \right) \left( a+d
\right) \left( {a}^{2}+a-{d}^{2}+2\,d-3 \right) }{ \left( a+d-1
\right) ^{2}}}
=0\;\;\;\;\;(2)$

From $(2)$ we see 4 possibilities: $d=-1,a=2$ or $a+d=0$. However, each one of these gives that $(1)$ cannot be satisfied so we are left with

${a}^{2}+a-{d}^{2}+2\,d-3=0$

together with

${a}^{2}d+{a}^{2}+a{d}^{2}-ad+2\,a-4+2\,d-2\,{d}^{2}=0$

Eliminating $d^2$ gives

$(a+2)(a-1)(a+d-1)=0$

which leads us to $a = 1$ and $a = -2$ from which we can find the remaining variables $b, c$ and $d$ leading to the solutions that Opalg found.

Thanks for participating, Jester! Your answer is correct and your solution is perfect too.:)

My solution:

[TABLE="class: grid, width: 800"]
[TR]
[TD]$a-b-c+d=1$ gives $a-b=1+c-d\;\;\;(1)$.

And squaring both sides of the equation $(a-b)^2=(1+c-d)^2$ yields $a^2+b^2-2ab=1+c^2+d^2-2cd+2c-2d\;(2)$.[/TD]
[TD]$a^2+b^2-c^2-d^2=3$ gives $a^2+b^2=3+c^2+d^2\;\;(3)$ .

Replacing it into (2) gives $3+c^2+d^2-2ab=1+c^2+d^2-2cd+2c-2d$ or

$2-2ab=-2cd+2c-2d$

$cd=c-d+ab-1$

$cd=a-b-1+ab-1=a-b+ab-2\;\;\;(4)$

$ab=cd-c+d+1\;\;\;(5)$[/TD]
[/TR]
[/TABLE]
$a^3-b^3-c^3+d^3=-5$ gives

$(a-b)(a^2+ab+b^2)-(c-d)(c^2+cd+d^2)=-5$

And from equations (1), (2) and (5), equation above becomes

$(1+c-d)(3+c^2+d^2+1-c+d+cd)-(c-d)(c^2+cd+d^2)=-5$

$9+3cd+3c-3d=0$

$cd=-c+d-3\;\;\;(6)$

Equations (1), (5) and (6) show us that $ab=2b-2a\;\;\;(7)$

On the other hand, if we want to keep the variables $a, b$, we see that we can also have

$(a-b)(a^2+ab+b^2)-(c-d)(c^2+cd+d^2)=-5$

$(a-b)(a^2+ab+b^2)-(a-b-1)(a-b+ab-2+a^2+b^2-3)=-5$

$4(a-b)+ab=-2\;\;\;(8)$

Equations (7) and (8) imply

$a-b=-1$ and therefore

$c-d=-3$, also,

$ab=-2-4(a-b)=-2-4(-1)=2$

Thus, we get

$a-b=-1$

$a-\frac{2}{a}=-1$

$a^2+a-2=0$

$(a+2)(a-1)=0$

That is, $a=-2$ or $a=1$.

The complete two solutions for the given system are therefore $(-2,-1,-1, 1)$ and $(1,2,-1,1)$.
 
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