How many ordered pairs of solutions (a,b) are there to a/b−b/a−2/a−2/b=0, where a and b are integers from −100≤a,b≤100?
To find the number of ordered pairs of solutions (a, b) that satisfy the given equation, we can approach it systematically by considering the possible values of a and b.
The equation is: (a/b) - (b/a) - (2/a) - (2/b) = 0
First, let's simplify the equation by finding a common denominator:
(a^2 - b^2 - 2b - 2a) / (ab) = 0
Next, multiply through by ab to eliminate the denominators:
a^2 - b^2 - 2b - 2a = 0
Rearrange the terms:
a^2 - 2a - b^2 - 2b = 0
Now, let's group the terms:
(a^2 - 2a) - (b^2 + 2b) = 0
We can see that this equation can be factored into two parts:
a(a - 2) - b(b + 2) = 0
Now, let's consider the two cases separately:
Case 1: a(a - 2) = 0
This implies that either a = 0 or a - 2 = 0, which gives us two possible values of a: a = 0 or a = 2.
Case 2: -b(b + 2) = 0
This implies that either b = 0 or b + 2 = 0, which gives us two possible values of b: b = 0 or b = -2.
Now, let's consider all possible combinations of these values for a and b:
(a, b) = (0, 0), (0, -2), (2, 0), (2, -2)
Therefore, there are four ordered pairs of solutions (a, b) that satisfy the given equation.
To summarize, the equation has four ordered pairs of solutions:
(0, 0), (0, -2), (2, 0), and (2, -2).