Quadratic Polynomials over $\mathbb F_p$ I'd like to know a reason why (irreducible) quadratic polynomials over $\mathbb F_p$ do not reach all numbers in $\mathbb F_p$.
Example:
$f(x)=x^2+3x+1$ in $\mathbb F_7$ is irreducible, i.e has no zeros but $f(\mathbb F_7)=\{1,4,5,6\}\neq \mathbb F_7$.
Kind regards,
reinbot
 A: This is just Robert Israel's (+1) answer rephrased.
The basic reason is that if $p$ is odd, then squaring, $x\mapsto x^2$, is (mostly) two-to-one as $(-x)^2=x^2$. The only violation is that $x=0$ is its own negative, so $0$ occurs as a value just once.
The general case is reduced to this by the familiar process of completing the square:
$$
x^2+ax+b=(x+\frac a2)^2+(b-\frac{a^2}4).
$$
The square part, $(x-\dfrac a2)^2$ has the same set of values with same frequency as the simpler $x^2$. The additive constant $b-a^2/4$ translates the set of values by that amount, but does not alter the frequency distribution, i.e. still 2-to-1 save a single exception.
Note that the assumption $2\nmid p$ was used as completing the square requires an ability to divide by two. We have this ability only in odd characteristic. Also note that irreducibility of the quadratic polynomial is irrelevant here. Irreducibility can be detected by checking whether zero is in the range or not, but the basic 2-to-1 property is always there. As the domain has an odd number of elements, we are forced to have that single exception to the 2-to-1 generic rule.
A: A quadratic polynomial over any field of odd characteristic is always two-to-one with only one exception, i.e. if $f(x) = y$ has a solution $r$ in $\mathbb F$, so that $x - r$ is a linear factor of $f(x) - y$, then $(f(x) - y)/(x-r)$ is another linear factor, giving another root in $\mathbb F$.  The two roots are distinct unless
$f'(r) = 0$.
