3. For a positive integer $k$, the equation
$$
\left(a^{2}-k\right)\left(b^{2}-k\right)=c^{2}-k
$$

has $\qquad$ integer solution sets $(a, b, c)$.
3. Infinitely many.

Let $a=b+1, c=ab-k$. Then
$$
\begin{array}{l}
c^{2}-k=a^{2} b^{2}-2 k a b+k^{2}-k, \\
\left(a^{2}-k\right)\left(b^{2}-k\right)=a^{2} b^{2}-k\left(a^{2}+b^{2}\right)+k^{2} .
\end{array}
$$

Since $a^{2}+b^{2}=(b+1)^{2}+b^{2}$
$$
=2 b(b+1)+1=2 a b+1 \text {, }
$$

Therefore, $\left(a^{2}-k\right)\left(b^{2}-k\right)=c^{2}-k$.
By the arbitrariness of $b$, the equation has infinitely many solutions.