6. All positive integer solutions $(a, b)$ of the equation $a^{3}-a^{2} b+a^{2}+2 a+2 b+1=0$ are $\qquad$ .
6. $(5,7)$.

From the given equation, it is easy to know that $a$ and $b$ are both odd numbers.
$$
\begin{array}{l}
\text { From } a^{3}-a^{2} b=a^{2}(a-b) \\
=-\left(a^{2}+2 a+2 b+1\right)<0 \\
\Rightarrow a<b .
\end{array}
$$

If $a=1$, then
$$
1-b+1+2+2 b+1=0 \Rightarrow b=-5 \text {. }
$$

Contradiction.
Therefore, $1<a<b$.
Thus, $b \geqslant a+2, a \geqslant 3$.
$$
\begin{array}{l}
\text { Hence } a^{3}+a^{2}+2 a+1=b\left(a^{2}-2\right) \\
\geqslant(a+2)\left(a^{2}-2\right) \\
\Rightarrow(a+1)(a-5) \leqslant 0 \\
\Rightarrow a \leqslant 5 .
\end{array}
$$

When $a=3$, $b=\frac{43}{7}$ (discard);
When $a=5$, $b=7$.
Therefore, the original equation has a unique positive integer solution
$$
(a, b)=(5,7) \text {. }
$$