Polynomials mod prime $p$ The problem is $5m^2+m+4 \equiv  0\pmod 7$.  I am supposed to first convert it to a quadratic whose first coefficient is $1$.  But the polynomial cannot be factored, so I am unsure as to how to do this.  Then, I am supposed to use the process of completing the square to convert it to a congruence of the form $y^2 \equiv a\pmod p$.
Can someone help me through this process?  
 A: Multiply by the inverse of $5$, which is $3$:
$$ 3 \times ( 5 m^2+m+4) \equiv 0 \pmod{7}, $$
which becomes
$$ m^2+3m+5 \equiv 0 \pmod{7} $$
Completing the square means finding $a$ and $b$ so the following is true
$$ m^2+3m+5 \equiv (m+a)^2+b \pmod{7} $$
Expanding the bracket,
$$ 3m+5 \equiv 2am + (a^2+b) \pmod{7} $$
The inverse of $2$ is $4$, so
$$ a =  4 \times 3 \equiv 5 \pmod{7}, $$
and
$$ 5 \equiv 5^2 +b \equiv 4+b \pmod{7}, $$
Hence $b \equiv 1 \pmod{7}$, and the answer is
$$ (m+5)^2 \equiv -1 \equiv 6 \pmod{7} $$
Now you just have to find the square roots of $6$. At this point, as @hardmath pointed out, we have a problem, since $6$ is not a square modulo $7$ (easy to check by computing them all, or there's some theorem about $-1$ being a square only if the prime is $1 \pmod{4}$, IIRC). Therefore there are no solutions.
A: $0 \equiv 5m^2\!+\!m\!+\!4\equiv -2m^2\!+8m\!+\!4 \iff 0\equiv m^2\!-\!4m\!-\!2 \equiv (m\!-\!2)^2+1\iff $  
$\!\!\iff (m\!-\!2)^2\equiv -1.\,$ But $\,-1\,$ is not a square: $\,j^2\equiv -1\,\overset{\rm cube}\Rightarrow\,  1\!\!\!\!  \underset{\overbrace{\rm Fermat^{\!\!\phantom{1}}}^{\phantom{.}}}\equiv \!\!\! j^6 \equiv (-1)^3\equiv -1$
