Find all positive composite integers $n$ greater than $1$ such that for any relatively prime divisors $a$ and $b$ of $n$ with $a > 1$ and $b > 1$, the number $ab-a-b+1$ is also a divisor of $n$.

Therefore: $n=k(ab-(a+b-1))$ but then I am not being able to conclude anything.

Any help will be truly appreciated.

  • $\begingroup$ What does the symbol 􀀀 mean? $\endgroup$ – Matthew Conroy Feb 10 '16 at 21:59
  • $\begingroup$ @MatthewConroy what symbol? $\endgroup$ – punctured dusk Feb 10 '16 at 22:12
  • $\begingroup$ @barto There appears to be an odd character added in for whitespace. I'll fix it. Cheers! $\endgroup$ – Matthew Conroy Feb 10 '16 at 22:25

Hint. If $a$ and $b$ are large enough, then $ab-a-b+1>\frac{ab}2$.
Indeed; the inequality is equivalent to $(a-2)(b-2)>2$, so it suffices that $a,b\geq3$ and $\{a,b\}\neq\{3,4\}$.

Prime powers are certainly solutions, as they don't have any pair of coprime divisors $>1$. Now suppose $n$ has two nontrivial coprime divisors whose product is $n$. By the above, there are two cases to consider: Case 1: at least one of them is $2$, which implies $n=2p^k$ for some odd prime $k$. We need $p^k-1\mid n$, so $p^k-1\mid2$ and hence $p^k=3$, so $n=6$. Case 2: $n=3\cdot4=12$. It's easy to check that this is a satisfies the property.
Summarizing: $n=p^k$ for some prime $p$ and an integer $k\geq0$, or $n=6$ or $n=12$.

  • $\begingroup$ this seems false. What about $n=91080$, $a=45$, $b=46$? In fact, for any $k$, let $n=k(k-1)(k-2)$, $a=k-1$, $b=k$, so that they divide $k$, and $ab-a-b+1=(a-1)(b-1)=(k-1)(k-2)$ divides $n$ $\endgroup$ – vrugtehagel Feb 10 '16 at 21:58
  • $\begingroup$ no we have to find n which satisfies this property. $\endgroup$ – jyoti prokash roy Feb 10 '16 at 22:01
  • $\begingroup$ @jyotiprokashroy, is that a response on my comment or on barto's answer? $\endgroup$ – vrugtehagel Feb 10 '16 at 22:02
  • $\begingroup$ @vrugtehagel I replied to you that the numbers which does not satisfy is not our concern. $\endgroup$ – jyoti prokash roy Feb 10 '16 at 22:04
  • $\begingroup$ @barto how come numbers which are power of a prime be in the solution set because thay dont have any relatively prime divisors. $\endgroup$ – jyoti prokash roy Feb 10 '16 at 22:05

HINT: $\space$ $ab-a-b+1=(a-1)(b-1)$ and two consecutive are always coprime (if $a=km$ and $a-1=kn$ then $1=k(m-n)$).

Hence if $a|n$ and $b|n$ then $ab(a-1)(b-1)|n$ however it is needed to consider also, for example $(b-2)$ and others. It follows a set of solutions $$\mathcal S=\{a!b!\space \text{where}\space (a,b)=1\}$$ and more generally $$\mathcal G=\{a_1!a_2!a_3!\cdot \cdot\cdot a_n!\space \text{for arbitrary natural numbers}\space a_i;i=1,2,3,…,n\}$$


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