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Let $p$ be a prime number, find all $n\in \mathbb Z^+$ such that $p+n\mid pn$.

Here is the source of the problem, go to the page number $155$.

First I want to note that $n=1$ is impossible, because $p+1>p$. I can’t solve this problem in general but for $p=2$, $n=2$ works and that’s the only solution because $$\frac{2n}{2+n}<2$$ Hence the only possible case is $2n=2+n$ which occurs at $n=2$

Edit : We can generalize this idea $$\lim_{n\to \infty } \frac{pn}{p+n} =p$$ And the function is increasing so we know $$\frac{pn}{p+n} <p$$ but that involves checking a lot of cases as $p$ gets bigger.

Bill Dubuque
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PNT
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  • Well, one cheap observation: writing $(p+n)\times k = pn$ we see that $nk=p(n-k)$ so $p,|,n$ or $p,|,k$. But your argument shows that $p,\nmid,k$ so we must have $p,|,n$. – lulu Jan 11 '22 at 20:28
  • I’ve just found that $n=p(p-1)$ works for all $p$ @lulu – PNT Jan 11 '22 at 20:39
  • Yes, those are all the solutions. I was just writing that up when I saw that @Mike settled the point. – lulu Jan 11 '22 at 20:41
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    Note that $p +n \lvert p^2 = p(p+n) - pn$. – TheOscillator Jan 11 '22 at 20:44

2 Answers2

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We now assume $p \ge 3$. Then $n \in \mathbb{N}$ must satisfy $n=(p-1)p$.

Indeed: First note that $p|n$, otherwise both $\gcd(n+p, p)$ and $\gcd(n+p,n)$ would be $1$, and thus $\gcd(n+p, np)$ would be $1$ as well. So write $n=kp$. Then if $n+p$ divides $np$, then $n+p=(1+k)p$ divides $np=kp^2$, which, as $p$ is prime, implies $1+k$ divides $kp$. However, $\gcd(1+k,k)=1$, so $1+k$ must divide $p$. As $p$ is prime however, it follows that $1+k$ must equal $p$, or equivalently, the equation $k=p-1$ must hold. So indeed, $n=kp$, with $k=p-1$.

Mike
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$\color{#c00}{\overbrace{a}^{\large p+\color{#0a0}n}}\!\mid p(\overbrace{\color{#c00}a\!-\!p}^{n})\!\iff \! a\mid p^2\!\!\iff\! a = p^2,\,$ by $\,\color{#c00}{a > p}\,$ by $\,\color{#0a0}n> 0.\ \ \bf\small QED$

Remark $ $ This is a special case of the easy general solution of $a+b\mid ab$.

Bill Dubuque
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