There's nothing specifying that the numbers have to be positive. In that case, there are an infinite number of solutions, or since we're talking computer, "billyuns and billyuns" of them as our favorite astronomer used to say. v = w = x = 3. y and z and arithmetic inverses of each other (1, -1), (2, -2), (3, -3), etc so their reciprocals add to zero and they drop out. That's four billion right there. There are 5 choose 2 = 10 ways of picking which pair of variables are arithmetic inverses, so make it 40 billion and I'm just getting warmed up. Shall we change the problem to requiring the integers to all be positive?
I guess we also have to specify that the formula is not computed with integer arithmetic. Otherwise, for any integer greater than 1, the division would yield 0, and thus, there would be "billyuns and billyuns" of possibilities too (2^160 for 32bit unsigned integers). ;)