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So my alarm clock makes me do some simple math problems to turn it off. This morning I confused $6 + 3$ for $6 \cdot 3$ but $18 = 2 \cdot 9$ which was slightly amusing to me. Solving $xy=N(x+y)$ for other $N$ (only the integer solutions of $x$ and $y$) using Wolfram Alpha gave an interesting pattern: whenever $N$ is a prime number, there are exactly $6$ unique integer solutions. When $N$ is not prime, the number of solutions is not exactly $6$.

Can this be proven?

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  • $\begingroup$ $N$ is usually reserved for $N=PQ$ where both $P$ and $Q$ are prime so it would be much better to post your question as: $xy = P(x+y)$. Just saying. $\endgroup$
    – steveK
    Feb 5 at 13:44

2 Answers 2

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$$xy=p(x+y)\Rightarrow xy-px-py+p^{2}=p^{2}\Rightarrow (x-p)(y-p)=p^{2}$$ We can split up $p^{2}$ into 2 factors in exactly 6 different ways ($p\cdot p$, $-p\cdot -p$, $1\cdot p^{2}$, $-1\cdot -p^{2}$, $p^{2}\cdot 1$, $-p^{2}\cdot-1$), giving rise to the 6 solution pairs.

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Alternately, $$xy = N(x+y)$$ can be solved for $y$ to give $y = Nx/(x-N)$. If you then do the polynomial division you can write

$$ y = N + {N^2 \over x-N} $$

and so as $x$ ranges over the integers, you get integer $y$ exactly when $x - N$ is a factor of $N^2$. So the number of solutions is just the number of factors of $N^2$, positive or negative.

If $N$ is prime then $N^2$ has three positive factors, namely $1, N, N^2$, and thus six total. But if $N$ is not prime then $N^2$ will have more than three positive factors and thus you'll have more than six solutions.

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