# Find triples $(a,b,c)$ of positive integers such that…

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