1. ## solutions of a biquadratic equation.

Consider the biquadratic equation, x^4 + x^3+x^2 +x +1 =0 in Z _11

I find that the equation has no solutions from a brute force method of trying all numbers from 1 to 10 for x.

I would like to know if there is another method to actually prove that there is no solution for this equation.

MAX

2. Originally Posted by MAX09
Consider the biquadratic equation, x^4 + x^3+x^2 +x +1 =0 in Z _11

I find that the equation has no solutions from a brute force method of trying all numbers from 1 to 10 for x.

I would like to know if there is another method to actually prove that there is no solution for this equation.

MAX
Surely x=3, 4, 5 and 9 are solutions.

Note that $x^4+x^3+x^2+x+1\equiv 0\ (mod\ 11)\Leftrightarrow \frac{x^5-1}{x-1}\equiv 0\ (mod\ 11)$

$x^{10}\equiv 1\ (mod\ 11)$ (Fermat's little theorem)

$\Rightarrow (x^5-1)(x^5+1)\equiv 0\ (mod\ 11)$

$\Rightarrow x^5\equiv \pm1\ (mod\ 11)$

etc.

3. Originally Posted by alexmahone
Surely x=3, 4, 5 and 9 are solutions.

Note that $x^4+x^3+x^2+x+1\equiv 0\ (mod\ 11)\Leftrightarrow \frac{x^5-1}{x-1}\equiv 0\ (mod\ 11)$

$x^{10}\equiv 1\ (mod\ 11)$ (Fermat's little theorem)

$\Rightarrow (x^5-1)(x^5+1)\equiv 0\ (mod\ 11)$

$\Rightarrow x^5\equiv \pm1\ (mod\ 11)$

etc.
I can't fault your logic, but I'm getting the solution to $x^5 \equiv \pm 1~\text{(mod 11)}$ to be all numbers x in Z_11! Obviously all of these cannot solve the original equation. I don't see how this gets us anywhere. How do you whittle them down to the solutions to the original problem?

-Dan

Edit: Oh Heavens I'm being silly. I got it. (I hope you aren't responding to this yet.)

4. Originally Posted by topsquark
I can't fault your logic, but I'm getting the solution to $x^5 \equiv \pm 1~\text{(mod 11)}$ to be all numbers x in Z_11! Obviously all of these cannot solve the original equation. I don't see how this gets us anywhere. How do you whittle them down to the solutions to the original problem?

-Dan
The four solutions (3, 4 ,5 and 9) satisfy $x^5\equiv 1\ (mod\ 11)$

On the other hand, the numbers 2, 6, 7, 8 and 10 satisfy $x^5\equiv -1\ (mod\ 11)$

5. Originally Posted by alexmahone
The four solutions (3, 4 ,5 and 9) satisfy $x^5\equiv 1\ (mod\ 11)$

On the other hand, the numbers 2, 6, 7 and 8 satisfy $x^5\equiv -1\ (mod\ 11)$
Yup. I got it (see my last edit), but apparently you were responding before you got the edit. Sorry about that!

-Dan