# Second Question about Reversing Equations

• Mar 24th 2011, 11:55 PM
haedious
Second Question about Reversing Equations
Here another similar question but more complicated than former:

$
\begin{cases}
& u=a^3+3b^2d+3bc^2+3ce^2+3d^2e+6abe+6acd \\
& v=b^3+3a^2d+3ae^2+3cd^2+3c^2e+6abc+6bde \\
& y=d^3+3a^2e+3ac^2+3b^2c+3be^2+6abd+6cde \\
\end{cases}$

I would interesting how many solutions when all variables are real numbers. And difference with when all variables are complex numbers.
• Mar 25th 2011, 06:08 AM
Opalg
Quote:

Originally Posted by haedious
Here another similar question but more complicated than former:

$
\begin{cases}
& u=a^3+3b^2d+3bc^2+3ce^2+3d^2e+6abe+6acd \\
& v=b^3+3a^2d+3ae^2+3cd^2+3c^2e+6abc+6bde \\
& y=d^3+3a^2e+3ac^2+3b^2c+3be^2+6abd+6cde \\
\end{cases}$

I would interesting how many solutions when all variables are real numbers. And difference with when all variables are complex numbers.

The solution in the complex case will be similar to the previous problem. Let $\omega = e^{2\pi i/5}$, and for k=0,1,2,3,4 let $\lambda_k = u+\omega^{-k}v + \omega^{-2k}x + \omega^{-3k}y + \omega^{-4k}z.$

Then $(a + \omega^{k}b + \omega^{2k}c + \omega^{3k}d + \omega^{4k}e)^3 = \lambda_k$, and it follows that

$\displaystyle a = \tfrac15\sum_{k=0}^5\lambda_k^{1/3},\quad b = \tfrac15\sum_{k=0}^5\omega^k\lambda_k^{1/3},\quad c = \tfrac15\sum_{k=0}^5\omega^{2k}\lambda_k^{1/3},\quad d = \tfrac15\sum_{k=0}^5\omega^{3k}\lambda_k^{1/3},\quad e = \tfrac15\sum_{k=0}^5\omega^{4k}\lambda_k^{1/3}.$

Each $\lambda_k$ has three complex cube roots, and there are five of them, giving a total of $3^5 = 243$ complex solutions. The number of real solutions is much more problematic. It's not clear to me whether there will be any at all.
• Mar 25th 2011, 05:23 PM
haedious
Quote:

Originally Posted by Opalg
The solution in the complex case will be similar to the previous problem. Let $\omega = e^{2\pi i/5}$, and for k=0,1,2,3,4 let $\lambda_k = u+\omega^{-k}v + \omega^{-2k}x + \omega^{-3k}y + \omega^{-4k}z.$

Then $(a + \omega^{k}b + \omega^{2k}c + \omega^{3k}d + \omega^{4k}e)^3 = \lambda_k$, and it follows that

$\displaystyle a = \tfrac15\sum_{k=0}^5\lambda_k^{1/3},\quad b = \tfrac15\sum_{k=0}^5\omega^k\lambda_k^{1/3},\quad c = \tfrac15\sum_{k=0}^5\omega^{2k}\lambda_k^{1/3},\quad d = \tfrac15\sum_{k=0}^5\omega^{3k}\lambda_k^{1/3},\quad e = \tfrac15\sum_{k=0}^5\omega^{4k}\lambda_k^{1/3}.$

Each $\lambda_k$ has three complex cube roots, and there are five of them, giving a total of $3^5 = 243$ complex solutions. The number of real solutions is much more problematic. It's not clear to me whether there will be any at all.

Thanks again! (Clapping)
Your answer make me more clear then previous question.
About the real solutions I'm thinking if $n^{1/4}$ will outcome 2 real numbers: $\sqrt{n},-\sqrt{n}$ and 2 other complexs $\sqrt{n}i,-\sqrt{n}i$. But I agree it could be more problematic for sort of this questions.