roots

**sidi** · May 30th 2009, 08:47 AM

How many roots has the polynomial $x^2 - 5$ v $Z<br /> 257?$
I know it can be 0,1 or 2.

thank you

**NonCommAlg** · May 30th 2009, 11:40 AM

Originally Posted by putnam120

I'm not sure I understand the question the way it is written. Are you trying to ask:
How many solutions does $x^2-5\equiv 0\bmod{257}$ have?

If so then the answer is $2$ since 257 is prime. (Note: If $p$ is prime, then $\mathbb{Z}_p$ is a field.)

actually there's no solution!

applying QRT gives us: $\left(\frac{5}{257} \right)=\left(\frac{257}{5} \right)=\left(\frac{2}{5} \right)=-\left(\frac{5}{2} \right)=-\left(\frac{1}{2} \right)=-1.$

by the way, questions like this one should be posted in the number theory subforum.

**putnam120** · May 30th 2009, 03:31 PM

Oh whoops totally forgot about that. I'll remove the previous post.

Thread: roots

LinkBack

Thread Tools

Display

roots

Similar Math Help Forum Discussions

roots

Roots & Imaginary Roots

roots

Roots ><

Given 2 imaginary roots find other 2 roots

Search Tags