若 $|1001-a|+\sqrt{a-1002}=a$, 则 $a-1001^{2}=$
解：根据算术平方根的非负性,

$\therefore a-1002 \geq 0$, 解得 $a \geq 1002$,

$\therefore|1001-a|+\sqrt{a-1002}=a-1001+\sqrt{a-1002}=a$, 即 $\sqrt{a-1002}=1001$,

解得 $a=1001^{2}+1002$,

$\therefore a-1001^{2}=1001^{2}+1002-1001^{2}=1002$.