Hi all
any help with the attached question would be wonderful
The discrimant is,
$\displaystyle 121-103 = 18$
Consider,
$\displaystyle \left\{ 18\cdot 1,18\cdot 2 , ... , 18\cdot 9 \right\}$
There are $\displaystyle 9$ such numbers so that the remainders exceede $\displaystyle 19/2$. Thus the Legendre symbol is $\displaystyle (-1)^9=-1$.
$\displaystyle 9x^2 - 11x + 3 = 0$ (mod 19)
I can't figure out how to use Euler's criterion on this (and I'm not even going to try Gauss' Lemma) but I can show that it has no solutions.
Complete the square:
$\displaystyle 9 \left ( x^2 - \frac{11}{9}x \right ) = -3$
$\displaystyle 9 \left ( x^2 - \frac{11}{9}x + \left ( \frac{11}{18} \right )^2 \right ) = -3 + 9 \left ( \frac{11}{18} \right )^2$
$\displaystyle 9 \left ( x - \frac{11}{18} \right ) ^2 = \frac{13}{36}$
$\displaystyle (18x - 11)^2 = 13$
Remember this is all mod 19. But 13 doesn't have a square root in mod 19. So this equation has no solution.
Edit: Okay, I guess you could simply use the quadratic formula and come up with the same result, which would be a bit faster.
-Dan
I understand that. I was being lazy with my notation. Obviously when I multiply a number by $\displaystyle \frac{1}{9}$ I am multiplying it by 17, the multiplicative inverse of 9 in modulo 19. (I can get away with this only because 19 is prime, thus $\displaystyle \mathbb{Z}_{19}$ is a field.)
However it doesn't matter much since my method didn't answer any of his questions. I just posted it because I didn't realize your post addressed the question (I didn't realize that until later, I thought it was in reference to his last question), and wanted to tell him the answer is that there is no solution to the equation so that he would know what to expect as an answer when he used the methods he was required to use.
-Dan