3. Solve the system in integers:

$$
\left\{\begin{array}{c}
a b c d=c+199 \\
51 a^{6}+5 c^{2}+20|b+d|=2016
\end{array}\right.
$$
Answer: (-1,2,1,-100); (-1,-2,1,100); (-1,100,1,-2);(-1,-100,1,2).

Solution: Transform the first equation to $c(a b d-1)=199$. The number 199 is prime, so c can take the values 1, -1, 199, -199. Note that the second equation cannot be satisfied when $c=+/-199$. Therefore, $c=1,-1$.

Consider the case $c=1$ (the case $c=-1$ is considered similarly).

$$
\left\{\begin{array}{c}
a b d=200 \\
51 a^{6}+20|b+d|=2011
\end{array}\right.
$$

Note that $51 \cdot 16=3264>2016$, so $a=0,1,-1$. But a cannot be zero, since 2011 is not divisible by 20. Therefore, $|b+d|=98$ and $b d= \pm 200$. When $a=1$, there are no solutions, but when $a=-1$, we get 4 solutions: $b=2, d=-100 ; b=-2, d=100 ; b=100, d=-2 ; b=-100, d=2$.