It is an olympiad problem. Find all ordered triples of positive integers(a,b,c), Such that 1/a+1/b+1/c=3/4
Till now i got only 1 solutions, but i expect there are more than that.
I brought 4 to the LHS and got 4/a+4/b+4/c=3, it is trivial though, and hence a=b=c=4. Hence from the above one, i got 1 solution. I expect there are many more but could not find it, ido not know what to do next. Please help!
Edit: i found there must be 25 solutions in all
If the smallest denominator is $4$, then they are all $4$, giving $\frac{1}{4} + \frac{1}{4} + \frac{1}{4}$.
If the smallest denominator is $3$, then $\frac{1}{b} + \frac{1}{c} = \frac{5}{12}$. Either $b$ or $c$ must be less than or equal to $4$. If $b=4$, $c=6$, giving $\frac{1}{3}+ \frac{1}{4} + \frac{1}{6}$. If $b=3$, $c=12$, giving $\frac{1}{3} + \frac{1}{3} + \frac{1}{12}$.
If the smallest denominator is $2$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{4}$. Either $b$ or $c$ must be less than or equal to $8$. This gives several more solutions: $$\{a,b,c\} = \{2,5,20\}, \{2,6,12\}, \{2,8,8\}.$$
From there, you just need to figure out how many orderings of each triplet there are.
Assume without loss of generality that $2\le a\le b\le c$. Then
$$
\frac{3}{a}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}.
$$ Hence $2\le a\le 4$. If $a=2$, then $\frac{ 1}{b}+\frac{1}{c}=\frac{1}{4}$. Since $\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}$, we get $2\le b\le 8$. By Exhaustive search, we get $(b,c)=(5,20),(6,12),(8,8)$.
If $a=3$, then $\frac{1}{b}+\frac{1}{c}=\frac{5}{12}$ and $3\le b\le 24/5$. In the same way, this gives $(b,c)=(3,12),(4,6)$.
If $a=4$, then $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$ and $4\le b\le 4$. It is only possible for $(b,c)=(4,4)$.
These are all solutions of the equation with $a\le b\le c$. And any solution $(a,b,c)$ is a permutation of one of these.
