How many ordered pairs of solutions (a, b) are there to \frac{a}{b} - \frac{b}{a} - \frac{2}{a} - \frac{2}{b} = 0, where a and b are integers from -100 \leq a,b \leq 100?
To find the number of ordered pairs of solutions (a, b) that satisfy the given equation, we need to solve the equation and count the valid solutions.
Let's start by simplifying the equation:
\(\frac{a}{b} - \frac{b}{a} - \frac{2}{a} - \frac{2}{b} = 0\)
To simplify fractions involving variables a and b, we can combine them over a common denominator. The common denominator in this case is \(ab\):
\(\frac{a^2 - b^2 - 2b - 2a}{ab} = 0\)
Now, let's rearrange the equation by multiplying both sides by \(ab\):
\(a^2 - b^2 - 2b - 2a = 0\)
Next, let's factor the left side of the equation:
\((a - b)(a + b) - 2(a + b) = 0\)
Notice that we have a common factor of (a + b):
\((a + b)(a - b - 2) = 0\)
Now, we have two cases to consider:
Case 1: \(a + b = 0\)
If \(a + b = 0\), then a = -b. Since a and b are integers between -100 and 100, there are 101 possible values for a and 101 possible values for b. Therefore, in this case, there are \(101 \times 101 = 10201\) ordered pairs (a, b).
Case 2: \(a - b - 2 = 0\)
If \(a - b - 2 = 0\), then a - b = 2. Again, there are 101 possible values for a and 101 possible values for b. Therefore, in this case, there are \(101 \times 101 = 10201\) ordered pairs (a, b).
Finally, we need to exclude any duplicate solutions. We can see that in both cases, when we consider negative values of a and b, we obtain duplicates. So, we exclude the cases where a < 0 and b < 0.
Therefore, the total number of ordered pairs (a, b) that satisfy the equation is:
\(2 \times (101 \times 101) - 101 = 20402 - 101 = 20201\)