1. Let $a$, $b$, $c$ be prime numbers, and satisfy $a^{5} \mid \left(b^{2}-c\right)$, where $b+c$ is a perfect square. Find the minimum value of $a b c$.
1. 1958.

When $(a, b, c)=(2,11,89)$,
$2^{5} I\left(11^{2}-89\right)$.
At this time, $11+89=100$ is a perfect square.
Thus, $a b c=2 \times 11 \times 89=1958$.
Below, we prove: $(a b c)_{\min }=1958$.
(1) When $a=2$, $32 \mid \left(b^{2}-c\right)$.
Since $b, c$ are prime numbers, $b^{2} \neq c$.
(i) When $b^{2}>c$, by
$b^{2} \equiv 1,9,17,25(\bmod 32)$,

we know the possible values of $c$ are
$17,41,73,89,97,113, \cdots$.
When $c=17,41,73$, it is verified that $a b c>1958$;
When $c=89$,
$(a b c)_{\min }=1958(b \geqslant \sqrt{89+32}=11)$;
When $c \geqslant 97, b=3,5,7$, there are no valid $(a, b, c)$.
(ii) When $b^{2}<c$, by
$32 \mid \left(c-b^{2}\right)$,
we know the possible values of $c$ are
$41,73,109,149,193, \cdots$.
When $c=41,73,109,149,193$, it is verified that $a b c>1958$;
When $c \geqslant 241$, $b \geqslant \sqrt{241-32}=14$.
Thus, $a b c>2 \times 14 \times 241>1958$.
(2) When $a \geqslant 3$, similarly, $b^{2} \neq c$.
Assume $a b c>1958$.
When $a=3$,
$a b c>3 b\left(c-b^{2}\right)$
$\Rightarrow 3 b\left(c-b^{2}\right)c$.
When $a \geqslant 11$, $a b c>11 \sqrt{11^{5}}>1958$.
When $a=7$, $a b>7 \sqrt{7^{3}}>907$.
Then $c=2, b>\sqrt{7^{5}}>128$.
Also, $\sqrt{b+c} \in \mathbf{Z}$, thus, $b \geqslant 169-2=167$.
At this time, $a b c \geqslant 2 \times 7 \times 167>1958$.
When $a=5$, $c \equiv b^{2} \equiv 1,9,0(\bmod 5)$.
If $5 \mid b^{2}$, then $25 \mid b^{2}$.
Thus, $c \equiv 1,4(\bmod 5) \Rightarrow c \geqslant 11$.
At this time, $a b c>5 \sqrt{5^{5}} \times 11>1958$.
When $a=3$, $a b>3 \sqrt{3^{5}}$, then $c \leqslant 41$.
Also, $c \equiv 1(\bmod 3)$, thus,
$c=7,13,19,31,37$.
If $c=7$, then $b<100, b \equiv 13,14(\bmod 27)$.
At this time, $b=13,41,67$, and $b+c$ are not perfect squares.
If $c=13$, then $b<50, b \equiv 11,16(\bmod 27)$.
At this time, $b=11,43$, and $b+c$ are not perfect squares.
If $c \geqslant 19$, then $b<34$.
Thus, $b^{2}<5 \times 243$, verification shows it does not hold.
In summary, $(a b c)_{\min }=1958$.