Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$

I found this on a local question paper, and I am unable to solve it.

Any help will be appreciated.

  • It's best to assume WLOG $a \ge b \ge c$ for these types of equations. – Chad Shin May 5 '16 at 5:41
  • @AndréNicolas Except that the obvious solution has biggest 3! – almagest May 5 '16 at 6:47
  • @almagest: I messed up on the wording. To reword, the smallest of $a,b,c$ is $\le 2$. So examine two cases, (i) smallest is $1$ and (ii) smallest is $2$. – André Nicolas May 5 '16 at 6:51

We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.

Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $b>2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.

Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2<2$ we must have $b<3$. That gives the solution already noted of $(2,2,3)$.

  • Thanks it will be a great help. – user333900 May 5 '16 at 7:42
  • There is also the "solution" $(1,2,\infty)$ :) – 6005 Aug 11 '16 at 0:09

Suppose $a\geq 3$, then $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\leq\left(1+\frac{1}{3}\right)^3=\frac{64}{27}<3$$ (note that $a\leq b\leq c$), a contradiction. Hence $a=1,2$.

If $a=1$, it comes to solve $(1+1/b)(1+1/c)=3/2$. The same trick shows $b<5$. Now one may simply list all possible values of $b$.
The case $a=2$ can be solved similarly.

Answer: $(a,b,c) \in \{(1,3,8),(1,4,5),(2,2,3)\}$

  • +1 although this somewhat follows the same reasoning as the accepted answer. – 6005 Aug 11 '16 at 0:08

Let $a=b=2$ and $c=3$. So you have $(3/2)(3/2)(4/3)=3$.

  • It seems to me that you did not prove that this is the only solution. – user228113 May 5 '16 at 9:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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