AIME 2010 I · 第 5 题

Solution 1

Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$, where equality must hold so $b = a - 1$ and $d = c - 1$. Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$.

Note: We can also find that $b=a-1$ in another way. We know

$$a^{2} - b^{2} + c^{2} - d^{2}=(a+b)+(c+d) \implies (a+b)(a-b)-(a+b)+(c+d)(c-d)-(c+d)=0 \implies (a+b)(a-b-1)+(c+d)(c-d-1)=0$$ Therefore, one of $(a+b)(a-b-1)$ and $(c+d)(c-d-1)$ must be $0,$ because they are both non-negative if $a \geq b + 1$ and $c \geq d + 1$, which is the initial condition. Clearly, $a+b \neq 0$ since then one would be positive and negative, or both would be zero. Therefore, $a-b-1=0$ so $a=b+1$. Similarly, we can deduce that $c=d+1.$

~Edited by mathwizard123123

Solution 2

Since $a+b$ must be greater than $1005$, it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\boxed{501}$ possible values. Hence the answer is 500+1=501

Solution 3 (Substitution of Differences)

Let $x = a-b$, $y = b-c$, and $z = c-d$. Since $a > b > c > d \ge 1$, then $x, y, z$ are positive integers.

Rewrite the second equation using $a^2 - b^2 = (a+b)(a-b)$ and $c^2 - d^2 = (c+d)(c-d)$:

$$x(a+b) + z(c+d) = 2010$$ Since $a+b+c+d = 2010$, we can substitute $c+d = 2010 - (a+b)$:

$$x(a+b) + z(2010 - (a+b)) = 2010 \implies (x-z)(a+b) = 2010(1-z)$$ Case 1: $z > 1$

If $z \ge 2$, then $1-z$ is negative. For the LHS to be negative, we need $z > x$. However, even at minimum values ($x=1, z=2, a+b=1007$), the magnitude of the LHS $|(1-2)1007|$ is far less than $|2010(1-2)|$.

Case 2: $z = 1$

If $z = 1$, the RHS becomes $0$. Since $a+b > 0$, we must have $x-z = 0 \implies x = 1$.

Thus, $a-b = 1$ and $c-d = 1$.

Finding the Range of $a$:

Substitute $b=a-1$ and $d=c-1$ into the sum:

$$2a + 2c - 2 = 2010 \implies a + c = 1006$$ Using the sequence $a > a-1 > c > c-1 \ge 1$:

$a-1 > c \implies a-1 > 1006 - a \implies 2a > 1007 \implies a \ge 504$ $c-1 \ge 1 \implies 1006 - a \ge 2 \implies a \le 1004$

Total values = 1004 - 504 + 1 = $\boxed{501}$

~LI,CHENXI