5. The positive integers $a, b, c, d$ satisfy $a>b>c>d, a+b+c+d=2010$ and $a^{2}-b^{2}+c^{2}-d^{2}=2010$. How many different sets of possible values of $(a, b, c, d)$ are there?
(1 mark)
正整數 $a$ 、 $b$ 、 $、 d$ 滿足 $a>b>c>d$ 、 $a+b+c+d=2010$ 及 $a^{2}-b^{2}+c^{2}-d^{2}=2010$ 。問 $(a, b, c, d)$ 有多少組不同的可能值?
(1 分)
5. 501
5. We have $a-b \geq 1$ and $c-d \geq 1$. As
$$
2010=a^{2}-b^{2}+c^{2}-d^{2}=(a+b)(a-b)+(c+d)(c-d) \geq(a+b)+(c+d)=2010,
$$
equality holds throughout and hence we must have $a-b=c-d=1$. This gives $(b+1)+b+(d+1)+d=2010$, or $b+d=1004$, where $b$ and $d$ are positive integers with $b \geq d+2$. Hence we get 501 choices for $(b, d)$, namely, $(1003,1),(1002,2), \ldots,(503,501)$, and hence 501 choices for $(a, b, c, d)$, namely, $(1004,1003,2,1),(1003,1002,3,2), \ldots,(504$, $503,502,501)$.