A divisibility problem without using induction or exhaustion 
Problem: Determine the values of $X$ such that  $24\mid(18+37X-6X^2-X^3)$ without using induction or exhaustion.

This question was motivated by the solution set $$[a\quad b\quad c\quad d]=\left[\frac34\quad\frac{37}{24}\quad-\frac14\quad-\frac1{24}\right]$$ in this question; that is, when is the value of $a+bX+cX^2+dX^3$ an integer?
It is easy to show that it holds for all odd integers $X$, after letting $X=2K+1$ and arriving at the expression $K(K-1)(2K+11)\equiv0\pmod6$.
For even integers, it is harder. Letting $X=2K$, we get $$18+148K-96K^2-64K^3\equiv0\pmod{24}$$ or that $$4K^3-K-9\equiv0\pmod{12}$$ and we can further simplify this to $$K(1-2K)(1+2K)\equiv3\pmod{12}.$$ How can we find for which $K$ this congruence holds without mathematical induction or proof by exhaustion?

It is necessary and sufficient that $18+37X-6X^2-X^3$ is divisble by $8$ and by $3$. Reducing mod $3$ yields
$$18+37X-6X^2-X^3\equiv X+2X^3\pmod{3},$$
where $X+2X^3\equiv-X(X+1)(X+2)\pmod{3}$, so this is satisfied for all $X$.
Reducing mod $8$ yields
$$18+37X-6X^2-X^3\equiv2+5X+2X^2+7X^3\pmod{8},$$
which is zero precisely when $X\equiv1,3,5,6,7\pmod{8}$. So the expression is divisible by $24$ if and only if $X$ is odd or $x\equiv6\pmod{8}$.

Another way to see this, of course after the fact, is that
$$18+37X-6X^2-X^3=24(2X+1)-(X+1)(X+2)(X+3),$$
so the left hand side is divisible by $24$ precisely when $(X+1)(X+2)(X+3)$ is. Divisibility by $3$ is guaranteed as it is a product of three consecutive integers. The product is divisible by $8$ whenever 


*

*either $X$ is odd, as then $X+1$ and $X+3$ are congruent to $0$ and $2$ modulo $4$, in some order,

*or $X+2$ is divisible by $8$.

