Find all elements $b\in\mathbb{Z}_{17}$ such that for at least one $x \in \mathbb{Z}_{17}$: $x^2 + 3x + b = 0$ Find all elements $b\in\mathbb{Z}_{17}$ such that for at least one $x \in \mathbb{Z}_{17}$:
$x^2 + 3x + b = 0$
Hi everyone, this one has me wondering. I'm not all that confident with second degree equations in rings as it is, but how do I deal with the term $b$? Any tips or whole solutions would be nice.
 A: Since $17$ is prime, $\mathbb{Z}_{17}$ is not just a ring, but a field.
Solving equations in a field, works the same all the time.
Also you could just go about and solve 
$x^2+3x+0=0$
$x^2+3x+1=0$
...
$x^2+3x+16=0$
$x_{1,2}=-\frac{3}{2}\pm\sqrt{\frac{9}{4}-b}$
Since there are no fractions in $\mathbb{Z}_{17}$ you have to ask what $2^{-1}$ and $4^{-1}$ is.
We search $y\in\mathbb{Z}_{17}$ with $2y=1$ Since $2\cdot 9=18=1\mod 17$ it is $2^{-1}=9$
The same for 4y=1 we get y=13, hence $4^{-1}=13$
$x_{1,2}=-3\cdot 9\pm\sqrt{9\cdot 13-b}$
$x_{1,2}=-27\pm\sqrt{117-b}$
We can reduce this mod 17 as well. But this is not needed. It just makes the calculation easier, in my opinion.
$x_{1,2}=7\pm\sqrt{15-b}$
So this just one solution for $b=15$, and two solutions if $15-b$ is a square in $\mathbb{Z}_{17}$.
Note, that you can not write $15-b<0$ like in $\mathbb{R}$, since there is no order $<$ on $\mathbb{Z}_{17}$. It is a field that can not be arranged. Like $\mathbb{C}$.
A: In ${\Bbb Z}_{17}$ we have
$$\eqalign{x^2+3x+b=0\quad
  &\Leftrightarrow\quad x^2-14x+b=0\cr
  &\Leftrightarrow\quad (x-7)^2-49+b=0\cr
  &\Leftrightarrow\quad b=49-(x-7)^2\cr
  &\Leftrightarrow\quad b=-2-(x-7)^2\ .\cr}$$
So $b$ must be $-2$ minus a square.  The squares in ${\Bbb Z}_{17}$ are
$$0^2=0\ ,\quad 1^2=16^2=1\ ,\quad 2^2=15^2=4$$
and so on.  So the values of $b$ are
$$-2-0=-2=15\ ,\quad -2-1=-3=14\ ,\quad -2-4=-6=11\ ,$$
and a few more which I'm sure you can find for yourself.
A: Well by brute force you cal list all $x^2 + 3x$ and then $b$ must simply be it additive inverse of those values.
$x^2 + 3x + b\equiv 0\mod 17$
$b \equiv -x^2 - 3x = -x(x+3) \mod 17$.
So just list them all:  $-0*3 = 0; -1*4= 13; -2*5=7.... etc.$
It's interesting as $-14*0=0$ so there will be fewer than $17$ distinct answers.
$(7 - i)(10-i) = 70 - 17i +i^2 \equiv 70 + 17i + (-i)^2 =(7 + i)(10+i)$ so you only have to check for $x=0... 7$. 
They are $0,-4,7,-1,6,-6,3,2$.  I don't really see any pattern in that to get any further insight how to solve without brute force. 
