5. If $\sqrt{a^{2}-1996}$ is an integer, then the minimum value of the integer $a$ is
5. Consider the case of natural number $a$: Let $\sqrt{a^{2}-1996} = m$, then $a^{2}-m^{2}=1996$, $(a+m)(a-m)=1996$. Since $a+m$ and $a-m$ are both even numbers, and $1996=2 \times 998$ is the only way to split it into the product of two even numbers, we have $a+m=998, a-m=2$, solving for $a=500, m=498$. Since the largest natural number satisfying $\sqrt{a^{2}-1996}$ being an integer is 500, the smallest integer satisfying $\sqrt{a^{2}-1996}$ being an integer is -500.