Modular arithmetic polynomial question Is it possible to find all polynomials of the form  $ an^2 + bn +c  $ where a,b, and c are integers and such that 
$$ a+b+c \equiv  31  \pmod{54}  $$
$$ 4a+2b+c \equiv  3  \pmod{54}  $$
$$ 9a+3b+c \equiv  11  \pmod{54}  $$

That you're making them coefficients of a polynomial is irrelevant: the problem is "solve this linear system of three equations in three variables".
That you're solving systems of equations modulo 54 rather than systems of equations of real numbers doesn't change things much: the only real difference is how to compute the inverse of a number, and that some non-zero numbers can fail to be invertible.
Solving the first equation gives
$$ c \equiv 31 - a - b $$
plugging into the other equations gives
$$ 3a + b \equiv -28 \equiv 26 $$
$$ 8a + 2b \equiv -20 \equiv 34 $$
Solving the first of these gives
$$ b \equiv 26 - 3a $$
plugging into the other equation gives
$$ 2a \equiv -18 \equiv 36 $$
Solving this is a little trickier: the solution space to
$$ gux \equiv gv \pmod{gw}$$
is the same as the solution space to
$$ ux \equiv v \pmod w$$
(Note if the equation was $gux \equiv v \pmod{gw}$ where $v$ is not divisible by $g$, then there would be no solutions for $x$)
So to solve
$$ 2a \equiv 36 \pmod{54} $$
you factor out the 2
$$ a \equiv 18 \pmod{27} $$
$$ a \equiv 18, 45 \pmod{54} $$
Now backsolving gives
$$ a \equiv 18 \pmod{54} \qquad b \equiv 26 \pmod{54} \qquad c \equiv 41 \pmod{54} $$
$$ a \equiv 45 \pmod{54} \qquad b \equiv 53 \pmod{54} \qquad c \equiv 41 \pmod{54} $$

Of course, I could have used Gaussian elimination and its variants instead. Or I could have solved for $a$ first, or any other variation.
The big thing I didn't demonstrate is the Chinese remainder theorem approach. $54 = 2 \cdot 3^3$; Often, it's easier to solve a problem modulo $2$ then solve the problem modulo $3^3$, then use the Chinese remainder theorem to combine the solutions to modulo $54$ than it is to solve it directly.
