Find the number of ordered triplets

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

• I assume these must be integers? – Matt Samuel Jan 19 at 13:52
• Don't forget solutions with negative terms, like $(1,-2,4)$ – lulu Jan 19 at 13:53
• These are positive integers – user636268 Jan 19 at 13:54
• Please edit the question to include all the requirements you have in mind. – lulu Jan 19 at 13:55
• Yeah you are right – user636268 Jan 19 at 13:55

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.

• Your solution is correct! – user636268 Jan 19 at 14:05

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.