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ThePerfectHacker · #30 November 18th, 2007, 07:32 PM

Here is solution to #1. We want $x+y=B, xy=A$ that means $x^2+y^2+2xy=B^2\implies x^2+y^2=B^2-2A$ . But then $(x-y)^2 = x^2 + y^2 - 2xy = B^2 - 2A - 2A = B^2 - 4A \implies x-y = \sqrt\pm {B^2-4A}$ . And the system of equations $x+y = B \mbox{ and }x-y = \pm \sqrt{B^2-4A}$ always has a solution provided $B^2-4A \geq 0 \implies B^2 \geq 4A$ .

So we need to count the number of ways we can get $B^2\geq 4A$ for all pairs $(A,B)$ . Note if $B\geq 90$ . Then the maximum value $4A$ is $4(2007) = 8028$ which is always true for $B\geq 90$ . We we just need to count all pairs $(A,B)$ where $B\leq 79$ .

If $B=1$ then there is no $A$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 0.

If $B=2$ then there is just $A=1$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 1.

If $B=3$ then there is just $A=1,2$ in the pair $(A,B)$ so that $B^2 \geq 4A$ . So the count is 2.

If $B=4$ then we can pick $A=1,2,3,4$ . So the count is 4.

If $B=5$ then we can pick $A=1,2,3,4,5,6$ . So the count is 6.

The question is whether we can find a pattern. Yes! It is based on looking at even and odd cases. Say $B$ is even so $B=2c$ then $B^2 \geq 4A\implies 4c^2\geq 4A\implies c^2 \geq A$ in that case $A=1,2,3,...,c^2$ . So the count is $c^2$ .

If $B$ is odd, so, $B=2c+1$ then $(2c+1)^2 \geq 4A \implies 4c^2+4c+1\geq 4A \implies c(c+1)+.25\geq A$ . Which means $A=1,2,3,...,c(c+1)$ . So the count is $c(c+1)$ .

Now if we write out the numbers as we did for $B=1,2,3,4,5$ out further we will get:
$0,1,2,4,6,9,12,16,...$
Where (alternatively) $0=0\cdot 1,2=1\cdot 1, 6=2\cdot 3,12=3\cdot 4,...$
And (again alternatively) $1=1^2,4=2^2,9=3^2,16=4^2,...$ .
This list continous until $B=78,79$ on $78$ the value of $c=78/2 = 39$ . And for $79$ is will be $39(40)$ .

If we split this sum into even terms in the sequence and odd terms in the sequence we get:
$\sum_{n=1}^{39}n^2 + \sum_{n=1}^{39}n(n+1) = \sum_{n=1}^{39}2n^2+n = \frac{39(40)(79)}{3}+\frac{39(40)}{2} = 41080+780=41860$ .
But this is for $B\leq 79$ (I just realized it should have been $B\leq 89$ but I am too lazy to change it now. You get the idea).

Now just find the number of pairs $(A,B)$ for $B\geq 80$ which is just $80(2007)$ and add to the answer.