11. (1 mark) Find the $2002^{\text {nd }}$ positive integer that is not the difference of two square integers.
(1 分) 求第 2002 個不能寫成兩個平方整數的差的正整數。
11. Note that $(n+1)^{2}-n^{2}=2 n+1$, hence every odd positive integer is the difference of squares of integers. Also $(n+2)^{2}-n^{2}=4 n+4=4(n+1)$, hence any integer divisible by 4 is the difference of squares of integers. Now for a number of the form $4 n+2$, suppose $4 n+2=x^{2}-y^{2}$, then $x$ and $y$ must be of the same parity, hence $x^{2}-y^{2}$ is divisible by 4 , while $4 n+2$ is not. So the positive integers that cannot be expressed as the difference of two square integers are exactly those of the form $4 n+2$. The $2002^{\text {th }}$ member is $4(2001)+2=\underline{\underline{8006}}$.