3. Let $a, b \in \mathbf{N}^{*},(a, b)=1$. Find the smallest positive integer $c_{0}$, such that for any $c \in \mathbf{N}^{*}, c \geqslant c_{0}$, the indeterminate equation $a x+b y=c$ has non-negative integer solutions.
3. If the equation $a x+b y=a b-a-b$ has a non-negative integer solution $\left(x_{0}, y_{0}\right)$, then $a\left(x_{0}+1\right)+b\left(y_{0}+1\right)=a b$, thus $a\left|b\left(y_{0}+1\right), b\right| a\left(x_{0}+1\right)$. Combining with $(a, b)=1$, we know $a\left|\left(y_{0}+1\right), b\right|\left(x_{0}+1\right)$, leading to $a\left(x_{0}+1\right)+b\left(y_{0}+1\right) \geqslant 2 a b$, a contradiction. On the other hand, let $c \in \mathbf{N}^{*}, c>a b-a-b$, then by the general solution form of the equation $a x+b y=c$, we know there exists a solution $\left(x_{0}, y_{0}\right)$, such that $0 \leqslant x_{0}a b-a-b-a(b-1)=-b$, the number $y_{0}>-1$, thus $y_{0} \geqslant 0$. Therefore, the original equation has a non-negative integer solution.
In summary, the required $c_{0}=a b-a-b+1$.