Polynomials and mod?

Let $f(x) = ax^3 + bx^2 +cx+d$ . Since $f(x)$ is a third-degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that $f(1) = f(5) = f(6)$ , and $f(2)=f(3)=f(7)$ ; otherwise, more bends would be required in the graph. Since only the absolute value of $f(0)$ is required, there is no loss of generalization by stating that $f(1) = 12$ , and $f(2) = -12$ . This provides the following system of equations:

$f(1)=a+b+c+d=12$

$f(2)=8a+4b+2c+d=-12$

$f(3)=27a+9b+3c+d=-12$

$f(5)=125a+25b+5c+d=12$

$f(6)=216a+36b+6c+d=12$

$f(7)=343a+49b+7c+d=-12$

Using any four of these functions as a system of equations yields $a = \frac{96}{6113}, b = \frac{146136}{6113}, c = -\frac{585792}{6113}, d = \frac{512916}{6113}$ . Using these values and plugging in $x=0$ gives $|f(0)| = 72$ .