1. Find the positive integer tuple $(a, b, c)$, such that $a^{2}+b+3=\left(b^{2}-c^{2}\right)^{2}$.
1. $(a, b, c)=(2,2,1)$.

Let $N=\left|b^{2}-c^{2}\right|$. Then
$$
\begin{array}{l}
N^{2}=a^{2}+b+3 \geqslant 1+1+3=5 \\
\Rightarrow N \geqslant 3 .
\end{array}
$$

Thus, $b \neq c$,
$$
\begin{array}{l}
N=|b-c||b+c| \geqslant(b+1) \times 1=b+1 \\
\Rightarrow b \leqslant N-1 .
\end{array}
$$

Since $a^{2}<N^{2}$, we have $a \leqslant N-1$.
Therefore,
$$
\begin{array}{l}
0=N^{2}-\left(a^{2}+b+3\right) \\
\geqslant N^{2}-(N-1)^{2}-b-3 \\
=2 N-b-4 \\
\geqslant N-3 \\
\Rightarrow N \leqslant 3 .
\end{array}
$$

Thus, $N=3$.
From the condition for equality in equation (1), we get
$$
a=N-1=2, b=N-1=2 \text {. }
$$

Then $9=N^{2}=\left|b^{2}-c^{2}\right|^{2}=\left(c^{2}-4\right)^{2}$ $\Rightarrow c^{2}=1$ or $c^{2}=7$.
Since $c$ is a positive integer, we have $c=1$. Therefore, $(a, b, c)=(2,2,1)$.