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Find all $c\in\mathbb Z^+$ for which $\exists a,b\in\mathbb Z^+, a\neq b$ with $\begin{cases}a+c\mid ab\\b+c\mid ab\end{cases}$

For those $c$, prove only finitely many $(a,b)$ exist.

My attempt:

Let $(a+c,b+c)=d$. Then $\color{RoyalBlue}{d\mid ab}$ and $\text{lcm}(a+c,b+c)\mid ab\iff \frac{(a+c)(b+c)}{d}\mid ab$.

So $d\neq 1$, because $(a+c)(b+c)\mid ab\,\Rightarrow ab\ge (a+c)(b+c)$, impossible.

$a\equiv -c\equiv b\,\Rightarrow\, a\equiv b$ mod $d$. Then $\color{RoyalBlue}{ab\equiv 0}\equiv a^2\equiv b^2\,\Rightarrow\, d\mid a^2,b^2$.

So $p\mid d\mid a+c,b+c,a^2,b^2\,\Rightarrow\, p\mid a,b,c$, and $(a,b,c)\ge 2$.

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  • $\begingroup$ How do you deduce $a^2 \equiv b^2 \equiv 0 \pmod{d}$ in your reasoning? $\endgroup$ – A.P. May 2 '15 at 14:19
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    $\begingroup$ @A.P. $a\equiv b\implies ab\equiv a^2$ $\endgroup$ – Hagen von Eitzen May 2 '15 at 14:20
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Your findings so far are fine, but not (yet) fully conclusive towards the main problem.

Let $$C=\{\,c\in\mathbb Z^+\mid \exists a,b\in\mathbb Z^+\colon a\ne b\land a+c| ab\land b+c| ab\,\}$$ be the set we are looking for. For $c\ge 2$ we can let $$\tag{$\star$} a=c^3+c^2-c,\qquad b=c(a-1)=c(c+1)(c^2-1)$$Then certainly $0<a<b$ and we have $$ab = c^2(c^2+c-1)(c+1)(c^2-1)=(a+c)(c^2+c-1)(c^2-1)$$ and $$ ab=ac(a-1)=(b+c)(a-1).$$ This shows $c\in C$ for $c\ge 2$. On the other hand $1\notin C$ for if $a+1\mid ab$ then $\gcd(a+1,a)=1$ implies $a+1\mid b$; likewise $b+1\mid ab$ implies $b+1\mid a$. Specifically $b\ge a+1$ and $a\ge b+1$, which is absurd. Therefore we have $$C=\mathbb Z^+\setminus\{1\}.$$


Remains to show that for each $c\in C$ there are only finitely many $(a,b)$. Empirically, it seems that the choice for $a,b$ made in $(\star)$ is the maximal possible choice. Showing that $\max\{a,b\}>c(c+1)(c^2-1)$ is really impossible would of course show finiteness.

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  • $\begingroup$ How do my findings imply $c\mid a,b$? I could see that if $c$ were prime, because $p\mid a,b,c\,\Rightarrow\, p\mid c\,\Rightarrow\,p=c$, but you later said we never used the assumption that $c$ is prime. $\endgroup$ – user236182 May 2 '15 at 15:45
  • $\begingroup$ @user236182 Alright, we did use it there, but with hindsight, once we have found a way to compute suitable $a,b$ from given $c$, the proof that $a+c\mid ab$ and $b+c\mid ab$ does not use that $c$ is prime. - Maybe I should have dropped that paragraph altogether and started right away with the display formula. But that would then feel like unmotivated and out of the blue, wouldn't it? $\endgroup$ – Hagen von Eitzen May 2 '15 at 15:57
  • $\begingroup$ @A.P. Duely clarified and uncluttered $\endgroup$ – Hagen von Eitzen May 2 '15 at 16:13
  • $\begingroup$ It is a decent solution of part 1 and we'll wait for someone to prove part 2. Thanks. $\endgroup$ – user236182 May 2 '15 at 16:20
  • $\begingroup$ Thanks. It is now considerably easier to understand your argument. $\endgroup$ – A.P. May 2 '15 at 18:37

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