# Count solutions to $xy=(x+y)R!$

Question: Is there anyway to find how many integer solutions there are to the formula $$1=\frac{x\cdot y}{(x+y)\cdot (R!)}$$ for any given $R$. Is there anyway to compute it and if so, compute it quickly.

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Do you mean $((x+y)\cdot R)!$ or $(x+y)\cdot (R!)$ ? –  Zev Chonoles Dec 24 '11 at 12:12
A good hint is at MO: mathoverflow.net/questions/84210/count-solutions-closed –  David Mitra Dec 24 '11 at 12:16

Yes, by @David Mitra's hint (mathoverflow.net/questions/84210/count-solutions-closed), the equation reduces to $(x-n)(y-n)=n^2$ for $n=R!$. Now if $n=\prod_1^r{p_k^{e_k}}$ is the unique factorization into the $r$ distinct primes $p_k \leq R$ (e.g. where $e_k=\sum_{i=0}^{\infty}[np_k^{-i}]$), then the number of (positive) divisors of $n^2=\prod{p_k^{2e_k}}$ is $\tau(n^2)=\prod{(1+2e_k)}$ since each positive divisor is of the form $u=\prod{p_k^{f_k}}$ where $0\leq f_k \leq 2e_k$ (so that the exponents $f_k$ are $r$-tuples of integer lattice points inside a $r$-dimensional rectangular solid), and for each such $u$ dividing $n^2$, we can take $v=n^2/u$ and write two unique integer solutions: $(x,y)=(n+u,n+v)$ and $(x,y)=(n-u,n-v)$. Thus, the total number of solutions is $$2\;\tau(R!^2) = 2\;\tau(n^2) = 2\;\prod_{k=1}^{r}(1+2e_k)$$

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