Solving for Real Solutions: Systems of Equations with Powers

[anemone]

Solve for real solutions of

$a-b-c+d=1$

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

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

$a^4+b^4-c^4-d^4=15$

 

[Opalg]

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]

 

[Mathwebster]

My solution

Spoiler

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.

 

[anemone]

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!

My solution

Spoiler

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:

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

$a^2+b^2-c^2-d^2=3$ 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 $2-2ab=-2cd+2c-2d$ $cd=c-d+ab-1$ $cd=a-b-1+ab-1=a-b+ab-2\;\;\;(4)$ $ab=cd-c+d+1\;\;\;(5)$

$a^3-b^3-c^3+d^3=-5$ 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 $(1,2,-1,1)$.

 

[CellCoree]

To solve this system of equations, we can use the substitution method. First, we can rearrange the first equation to solve for $a$:

$a=1+b+c-d$

Then, we can substitute this value of $a$ into the other equations:

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

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

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

Expanding and simplifying these equations, we get:

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

$b^3+c^3-d^3+3b^2+3c^2-3d^2+3b+3c-3d=4$

$b^4+c^4+d^4+4b^3+4c^3+4d^3+6b^2+6c^2+6d^2+4b+4c+4d=16$

We now have a system of three equations with three variables. We can solve for $b$, $c$, and $d$ by using substitution and elimination methods. Once we have values for these variables, we can plug them back into the equation for $a$ to find its value.

It is important to note that there may be multiple solutions for this system of equations. We can check our solutions by substituting them into the original equations to see if they satisfy all four equations.