Problem (Number Theory):
Determine all integers $k$, with $k \neq 0$, for which the parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts.

Solution:
The parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts exactly when the discriminant of the quadratic equation $k x^{2}+6 x+k=0$ is positive.

Here, the disciminant equals $\Delta=6^{2}-4 \cdot k \cdot k=36-4 k^{2}$.

The inequality $36-4 k^{2}>0$ is equivalent to $k^{2}<9$.

Since $k$ is an integer and $k \neq 0$, then $k$ can equal $-2,-1,1,2$.

(If $k \geq 3$ or $k \leq-3$, we get $k^{2} \geq 9$ so no values of $k$ in these ranges give the desired result.)

Final answer: -2,-1,1,2