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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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
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
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