1. If $a$ and $b$ are both prime numbers, and $a^{2}+b=2003$, then the value of $a+b$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002
$-、 1 .(\mathrm{C})$.
Since 2003 is an odd number, one of $a$ or $b$ must be the even prime number 2. If $b=2$, then $a^{2}=2001$, but 2001 is not a perfect square. Therefore, $a=2, b=1999$, and we have $a+b=2001$.