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If a number can be expressed as a product of n unique primes, in how many ways can the number be expressed as a difference of two squares?

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    I was close to Cofopuff's solution but was stuck in the combinations......2012-08-09

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Write $m = p_1 \times \ldots \times p_n$ and for now assume each $p_i$ is odd. Then each factorization of $m = a\times b$ into a pair of odd numbers $(a,b)$ gives a representation $m = ab = \left(\frac{a+b}{2} + \frac{a-b}{2}\right)\left(\frac{a+b}{2} - \frac{a-b}{2}\right) = \left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$ and vice versa. So you need to consider all possible pairs of factors $(a,b)$ (where order doesn't matter). There are $\frac{1}{2}\left(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n}\right) = 2^{n-1}$ of these.

If one of the $p_i$ is two, then there are no representations: by the unique primes assumptions, $m \equiv 2 \bmod 4$ and all differences of two squares are one of $0,1,3 \bmod 4$.

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    @Swapnanil See also the [*composition law* for differences of squares.](http://math.stackexchange.com/a/97903/242)2012-08-09