smc

For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1,a_2,a_3,a_4).$ Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac{1}{m}$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$ Since $\gcd (m,4)=1,$ we conclude that $$\{k+4(0),k+4(1),k+4(2),\dots ,k+4(m-1)\}$$ is the complete residue system modulo $m$ for all integers $k.$ Given any ordered quadruple $(a_1^{\prime },a_2^{\prime },a_3^{\prime },a_4^{\prime })$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ \begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array} We conclude that $q(m)=\frac{1}{m}\cdot [m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so $$q(x)=(x-1)(x-2)(x-3)=c_3{x}^3+c_2{x}^2+c_{1}x+c_0.$$ By Vieta's Formulas, we get $c_1=1\cdot 2+1\cdot 3+2\cdot 3=\boxed{11}.$