3. Given the equation in $x$
$$
x^{2}+(a-2010) x+a=0 \quad (a \neq 0)
$$

has two integer roots. Then the value of the real number $a$ is $\qquad$.
3.4024.

Let the roots of the equation be $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$.
By Vieta's formulas, we have
$$
x_{1}+x_{2}=-(a-2010), x_{1} x_{2}=a \text {. }
$$

Then $x_{1} x_{2}+x_{1}+x_{2}=2010$, which means
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=2011 \text {. }
$$

Since 2011 is a prime number, we have
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = 0 , } \\
{ x _ { 2 } = 2 0 1 0 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=-2012, \\
x_{2}=-2 .
\end{array}\right.\right.
$$

Thus, $a=0$ (discard) or $a=4024$.