Answer:
The page numbers are 161, 162, 48197, and 48198.
Explanation:
Let the page numbers of the first facing pages be x and x+1, and let the page numbers of the second facing pages be y and y+1.
Then, we have two equations based on the information given in the problem:
x + (x+1) + y + (y+1) = 2x + 2y + 2 = 2(x+y+1)
(x + (x+1))(y+(y+1)) = (2x+1)(2y+1) = 4xy + 2x + 2y + 1
Multiplying these two equations, we get:
2(x+y+1)(4xy+2x+2y+1) = 96,717
Simplifying this expression, we get:
(2x^2+2y^2+5xy+3x+3y+1)(x+y+1) = 48,359
We can see that $48,359$ is a prime number, so it must have two factors: 1 and 48,359 or -1 and -48,359. Since x and y are positive integers, we can discard the negative solutions.
So we have two possibilities:
Case 1: x+y+1 = 1 and 2x^2+2y^2+5xy+3x+3y+1 = 96,717
In this case, x+y=0, which is not possible since x and y are positive integers.
Case 2: x+y+1 = 48,359$ and $2x^2+2y^2+5xy+3x+3y+1 = 2
In this case, x+y=48,358. We can solve the second equation for y in terms of x:
y = \frac{-5x-3\pm\sqrt{25x^2+30x-23}}{4}
Since $y$ must also be a positive integer, we can see that the only solution is y = \frac{-5x-3+\sqrt{25x^2+30x-23}}{4}. We can substitute this into the equation x+y=48,358 to get:
x+\frac{-5x-3+\sqrt{25x^2+30x-23}}{4} = 48,358
Simplifying this equation, we get:
29x^2+7x-185,287=0
Using the quadratic formula, we get:
x = \frac{-7\pm\sqrt{185,321}}{58}
We discard the negative solution, so:
x = \frac{-7+\sqrt{185,321}}{58} \approx 160.71
Since x must be a positive integer, the closest integer solution is x=161.
Substituting this into the equation x+y=48,358, we get:
y = 48,358 - 161 = 48,197
Therefore, the page numbers are 161, 162, 48197, and 48198.