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 $(a,b,c)\ge 2$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @A.P. $a\equiv b\implies ab\equiv a^2$ $\endgroup$ – Hagen von Eitzen May 2 '15 at 14:20

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.