Number of integer solutions (ordered and unordered)

Question

$$\frac1 a + \frac 1 b +\frac 1 c = \frac 34$$

Find number of triplets of $a\ , b\ , c\in \mathbb{Z}^+$

Should it not be infinite since it can be $\frac 34$ or $\frac38$ or $\frac9{12}$ etc. till infinity?

Please help since I am clueless on how to even begin solving this.

Answer 1

Probably the most elementary way to do this is bounding the variables. W.L.O.G. suppose that $a \le b \le c$. Observe that if $a \ge 5$ then $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \le \frac{3}{5}<\frac{3}{4}$$Thus $a \le 4$. $a=1$ doesn't work. Hence we have three cases to check: $a=2, 3, 4$. If $a=2$ equation reduces to$$\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$$Obviously $b>4$ and if $b>8$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{9} < \frac{1}{4} $$which is again a contradiction. So $ 5\le b \le 8$ and from here you get solutions $(a, b, c)=(2, 5, 20), (2, 6, 12), (2, 8, 8)$. For $a=3$ you get$$\frac{1}{b}+\frac{1}{c}=\frac{5}{12}$$Here we must have $b>2$ and if $b>4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{5} < \frac{5}{12} $$So $ 3\le b \le 4$ and from here you get solutions $(a, b, c)=(3, 3, 12), (3, 4, 6)$. For $a=4$ you get$$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$$Here we must have $b>3$ and if $b>4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{5} < \frac{1}{2} $$Hence you must have $ b=4$, which gives one solution $(a, b, c)=(4, 4, 4)$.

Answer 2

Write this as the following: $$\frac 1 a+\frac 1 b+\frac 1 c=\frac{3n}{4n}$$ Then multiply both sides by $4n$: $$\frac {4n} a+\frac {4n} b+\frac {4n} c=3n$$ Now, we need to find three divisors of $4n$ such that they sum to $3n$.

• $n=1 \implies$ Obviously, we have $1+1+1=3$. This gives us $a=b=c=\frac 4 1=4$.
• $n=2 \implies$ The divisors of $8$ are $1,2,4,8$, so we have $6=2+2+2$. This gives us $a=b=c=\frac 8 2=4$. (From now on, we know that all of the sums that are just $x+x+x$ will just give us $a=b=c=4$, so we're going to ignore that sum.) We also have $6=1+1+4$ which gives us $a=\frac 8 1=8$, $b=\frac 8 1=8$, and $c=\frac 8 4=2$.
• $n=3 \implies$ The divisors of $12$ are $1,2,3,4,6,12$. We can sum to $9$ with $1+2+6$, so we get $a=\frac {12} 1=12$, $b=\frac{12}{2}=6$ and $c=\frac{12}{6}=2$. We also have $9=2+3+4$, giving us $a=\frac{12}{2}=6$, $b=\frac{12}{3}=4$ and $c=\frac{12}{4}=3$. Furthermore, we have $9=1+4+4$, giving us $a=\frac{12}{1}=12$, $b=\frac{12}{4}=3$, and $c=\frac{12}{4}=3$

Keep going to find more solutions. Just for recap, here are my five solutions for $(a, b, c)$: $$(4, 4, 4)$$ $$(8, 8, 2)$$ $$(12, 6, 2)$$ $$(6, 4, 3)$$ $$(12, 3, 3)$$ However, there is still one more solution. Can you find it?

Answer 3

We start by looking for unordered solutions (ie we regard $(a,b,c)=(2,8,8)$ and $(8,2,8)$ as the same solution.

Suppose $a=b=c$, then clearly there is the unique solution $(4,4,4)$.

Suppose $a=b\ne c$. We have $8c+4a=3ac$, so $(3a-8)c=4a$. The LHS must be positive, so $a\ge 3$. If $a>8$, then $3a-8>2a$ and $3a-8<4a$, so $c=1$ is too small and $c\ge2$ is too big. That leaves $a=3,4,5,6,7,8$. It is easy to check that 3 and 8 give the solutions $(3,3,12),(8,8,2)$.

So it remains to consider the case $a<b<c$. We have $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}<\frac{45}{60}=\frac{3}{4}$. So we must have $a\le3$. We cannot have $a=1$ because $1>\frac{3}{4}$. So $a=2$ or 3. Consider first $a=3$.

We have $\frac{1}{3}+\frac{1}{5}+\frac{1}{6}=\frac{7}{10}<\frac{3}{4}$. So we must have $b=4$. It is easy to check this gives the solutions $(3,4,6)$ and no others.

Finally consider $a=2$. We have already know that $(2,8,8)$ is a solution, so any solution with $a<b<c$ must have $b<8$. It must have $b>4$ since $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$. So we check $b=5,6$ and find the solutions $(2,5,20),(2,6,12)$.

That gives a total of 6 unordered solutions. Three have $a,b,c$ unequal, so each generate 6 ordered solutions. Two have just two of $a,b,c$ equal, so they each generate 3 ordered solutions. Finally, the solution with $a=b=c$ gives only one ordered solution, for a total of 25 ordered solutions.