If p is the difference between a number and its reciprocal and q is the difference between the square of the same number and the square of its reciprocal, then what is p^{4}+4p^{2} equal to?

Explanation

Let the number be x. Given p = x - \frac{1}{x} and q = x^2 - \frac{1}{x^2}. We need p^4 + 4p^2 = p^2(p^2+4). Since p^2 = (x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}, we have p^2+4 = x^2 + 2 + \frac{1}{x^2} = (x + \frac{1}{x})^2. Multiplying them: p^2(p^2+4) = (x - \frac{1}{x})^2 (x + \frac{1}{x})^2 = ((x - \frac{1}{x})(x + \frac{1}{x}))^2 = (x^2 - \frac{1}{x^2})^2 = q^2.

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