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Let $p$ be a prime number. Find all positive integers $x$ and $y$ for which $$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{p}.$$

Multiplying the given expression by $xy$ results in $y+x = \dfrac{xy}{p} \Rightarrow p(x+y) = xy$.

I was suggested to get this to the form $(x-p)(y-p) = p^2$, which seems a bit weird to me since I usually with these kinds of questions I would find a system which would lead to finding the correct $(x, y)$, but having the term $p^2$ is not something I've seen before.

Is the idea behind this so that I would want to have an expression where I have something of the form $(x-k)(y-n) = p$ and could deduce from here that either $(x-k) = 1$, $(y-n) = p$ or the other way around since $p$ can only have the factors $1$ and $p$?

StubbornAtom
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2 Answers2

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Once you get to $(x-p)(y-p)=p^2$ you use unique factorisation of integer to state that either $x-p=y-p=p$ leading to $x=y=2p$ or $x-p=p^2$ and $y-p=1$ leading to $x=p^2+p$ and $y=p+1$

The case $x-p=-p^2$ and $y-p=-1$ is to be excluded because $x$ and $y$ are assumed to be positive integers

marwalix
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    In general if $p$ is prime, the only divisors of $p^n$ are the powers of $p$ up to $p^n$ ($p^0, p^1, p^2 \cdots p^n)$ since $p$ has no other divisors other than $1$ and itself. – Toby Mak Jul 18 '20 at 10:09
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$$p(x+y)=xy$$

$$p^2=xy-px-py+p^2$$

$$p^2=(x-p)(y-p)$$

Once you reach this form, we can consider a few cases:

Case $1$: $x-p = p$, $y-p=p$. That is $x=2p$ and $y=2p$.

Case $2$: $x-p=-p$, $y-p=-p$, which means $x=0$ which is not what we want.

Case $3$: $x-p=p^2$, $y-p=1$. $x=p(p+1)$, $y=p+1$.

Case $4$: $x-p=1$, $y-p=p^2$. Similar as case $3$.

Case $5$: $x-p=-p^2$, of which $x <0$, which is not what we want.

Similarly if $y-p=-p^2$.

Siong Thye Goh
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