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- Feb 14, 2012

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$a-b-c+d=1$

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

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

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

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- #1

- Feb 14, 2012

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$a-b-c+d=1$

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

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

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

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- Feb 7, 2012

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The only way I can do this is by guesswork.

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

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

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

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

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.

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

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

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

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?

Last edited:

- Jan 26, 2012

- 183

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.

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- Feb 14, 2012

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Thanks for participating,The only way I can do this is by guesswork.

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

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

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

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.

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,

Thanks for participating,

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.

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 $(1,2,-1,1)$.