1. ## Legendre Symbol

This question has completely floored me.
__________________________________

Let $f(x) = x^7 -2x^6-2x^5-3x^4$. Evaluate the Legendre symbol

$\displaystyle \left( \frac{f(x)}{19} \right)$

for all integers x.
__________________________________

I have factorised f(x) and have the following, by multiplicativity:

$\displaystyle \left( \frac{f(x)}{19} \right) = \left( \frac{x}{19} \right)^4 \left( \frac{x-3}{19} \right) \left( \frac{x^2 + x +1}{19} \right)$

and the first term must be +1 (as it is raised to an even power). But I've been manipulating the other two for the last two hours and have gotten pretty much nowhere. Any ideas?

2. Originally Posted by Capillarian
This question has completely floored me.
__________________________________

Let $f(x) = x^7 -2x^6-2x^5-3x^4$. Evaluate the Legendre symbol

$\displaystyle \left( \frac{f(x)}{19} \right)$

for all integers x.
__________________________________

I have factorised f(x) and have the following, by multiplicativity:

$\displaystyle \left( \frac{f(x)}{19} \right) = \left( \frac{x}{19} \right)^4 \left( \frac{x-3}{19} \right) \left( \frac{x^2 + x +1}{19} \right)$

and the first term must be +1 (as it is raised to an even power). But I've been manipulating the other two for the last two hours and have gotten pretty much nowhere. Any ideas?

$x^2+x+1=(x+8)(x+12)\!\!\pmod{19}$ , so taking into account your first deduction we get

$\left(\frac{f(x)}{19}\right)=\left(\frac{x+16}{19} \right)\left(\frac{x+8}{19}\right)\left(\frac{x+12 }{19}\right)$.

Now input all the values from $0\,\,to\,\,18\!\!\pmod{19}$ , and with a list of the squares modulo 19 by your side check each output...

Tonio

3. Edit: Terrible formatting. See below.

4. Wonderful! Thanks so much. That question was bugging me for ages. I got the following:

$\displaystyle \left( \frac{f(x)}{19} \right) = \begin{Bmatrix} 0 & \text{ if } x \equiv 0, 4, 7, \text{ or } 11 (\mod 19) \\ +1 & \text{ if } x \equiv 2, 3, 8, 10, 14, \text{ or } 17 (\mod 19) \\ -1 & \text{ otherwise. } \end{matrix}$

Does that seem the right sort of thing?

5. Originally Posted by Capillarian
Wonderful! Thanks so much. That question was bugging me for ages. I got the following:

$\displaystyle \left( \frac{f(x)}{19} \right) = \begin{Bmatrix} 0 & \text{ if } x \equiv 0, 4, 7, \text{ or } 11 (\mod 19) \\ +1 & \text{ if } x \equiv 2, 3, 8, 10, 14, \text{ or } 17 (\mod 19) \\ -1 & \text{ otherwise. } \end{matrix}$

Does that seem the right sort of thing?

Why do you get zero with $4\!\!\pmod {19}$ ? I get 1...Likewise, with $2\!\!\pmod {19}$ I get $-1$ ...Check all these.

Tonio

6. Originally Posted by tonio
Why do you get zero with $4\!\!\pmod {19}$ ? I get 1...Likewise, with $2\!\!\pmod {19}$ I get $-1$ ...Check all these.

Tonio
Argh! I've accidentally written the factorization down as $f(x) = (x+8)(x+12)(x+15) \mod 19$ when the last term should have been 16. That's it -- I get the same answer as you now. Thanks again!