[Number Theory]
Problem: Suppose that $a=5$ and $b=4$. Determine all pairs of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.

Solution: When $a=5$ and $b=4$, we obtain $a^{2}+b^{2}-a b=5^{2}+4^{2}-5 \cdot 4=21$.

Therefore, we want to find all pairs of integers $(K, L)$ with $K^{2}+3 L^{2}=21$.

If $L=0$, then $L^{2}=0$, which gives $K^{2}=21$ which has no integer solutions.

If $L= \pm 1$, then $L^{2}=1$, which gives $K^{2}=18$ which has no integer solutions.

If $L= \pm 2$, then $L^{2}=4$, which gives $K^{2}=9$ which gives $K= \pm 3$.

If $L= \pm 3$, then $L^{2}=9$. Since $3 L^{2}=27>21$, then there are no real solutions for $K$.

Similarly, if $L^{2}>9$, there are no real solutions for $K$.

Therefore, the solutions are $(K, L)=(3,2),(-3,2),(3,-2),(-3,-2)$.

Final answer: $(3,2),(-3,2),(3,-2),(-3,-2)$