So the question asks to solve for real valued $a$ such that $b,c,d\in\mathbb{R}$ $$abcd=-1$$ $$(a+c)(b+d)=-1$$ $$ac+bd+a+b+c+d=-1$$ $$ab+cd=ac+a+c$$ So assuming the four numbers are roots of a quartic equation $x^4+b_3x^3+b_2x^2+b_1x+b_0=0$. How do we use Vieta relation or some other method to solve this ?

I could not get all the $b_i$ using the above equations without solving $a,b,c,d$ indivdually. Any ideas? Thanks

  • $\begingroup$ Note that the second equality can be used in the third one to get the sum of the roots of your equation. $\endgroup$ – DiegoMath Oct 24 '14 at 3:33
  • $\begingroup$ Yes but no. There are extra terms $\endgroup$ – amathnerd Oct 24 '14 at 4:10

Curiously, this system turns out to involve $x^{17}-1=0$.

You have four equations in four unknowns. Resolving this to a single equation, we get two non-trivial octics, one with all real roots, the other with all complex roots. The relevant roots are,

$$\begin{aligned} a_k &=2\cos\Bigl(\frac{2k\,\pi}{17}\Bigr)\\ b_k &=2\cos\Bigl(\frac{4k\,\pi}{17}\Bigr)\\ c_k &=2\cos\Bigl(\frac{8k\,\pi}{17}\Bigr)\\ d_k &=2\cos\Bigl(\frac{16k\,\pi}{17}\Bigr) \end{aligned}$$

for $k = 1,2,3,\dots 8$. As a quartic, all these have the same form, namely,

$$-1 + (2 \mp \sqrt{17}) a + \tfrac{1}{2} (-3 \pm \sqrt{17}) a^2 + \tfrac{1}{2} (1 \pm \sqrt{17}) a^3 + a^4=0\tag1$$

and similarly for $b,c,d$.

P.S. It seems doubtful one can exclusively find $(1)$ without solving for $a,b,c,d$ individually, considering there is another quartic which yields valid but non-real solutions.


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