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I am demanded to figure out the number of integer solutions of this equation:

$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$

where $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$, $p_i$ being its prime factors.

I already got the hint that the above equation can be written as $(x-n)(y-n)=n^2$, however I am still stack.

What should be my next step?

Mr.Infinity
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  • Clarify: what are given (specific n)? Are x and y integers? Equation simplifies to xy=n(x+y). – herb steinberg May 29 '21 at 20:54
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    For each divisor of $n^2$ you get a value(s) that $x-n$ could take. You can compute the number of divisors of $n^2$ this way. Look at $\sigma_0(n^2)$. You might need consider $\pm$ each divisor, or perhaps exclude some divisors, depending on whether your solutions should be integers, or only positive integers. – plop May 29 '21 at 20:55

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Let $x=n+e.$ Then $y=n+n^2/e.$ So $e=\pm d$ where $d$ is a positive divisor of $n.$ So it is $necessary$ that $(x,y)$ is

Type 1: $(x,y)=(n+d,n+n^2/d)$ with $0<d|n$

OR Type 2: $(x,y)=(n-d,n-n^2/d)$ with $0<d|n^2.$

It is easily seen that either Type is also $sufficient.$

A Type 1 pair $(x,y)$ cannot be a Type 2 pair $(x',y'),$ as $x>n>x'.$

If $0<d|n$ and $0<d'|n$ we cannot have $(n+d, n+n^2/d)=(n+d', n+n^2/d')$ nor $(n-d,n-n^2/d)=(n-d', n-n^2/d')$ unless $d=d'.$

So for each positive divisor $d$ of $n^2$ we have $2$ solutions, adding up to $2\delta (n^2)$ solutions, where $\delta (n^2)$ is the number of positive divisors of $n^2.$

If $|n|=1$ then $\delta (n^2) =1.$

If $|n|>1$ and if $\prod_{i=1}^kp_i^{\alpha_i}$ is the prime decomposition of $|n|$ then $n^2=\prod_{i=1}^kp_i^{2\alpha_i},$ so $$\delta(n^2)=\prod_{i=1}^k(1+2\alpha_i).$$

DanielWainfleet
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