solve in positive integers $\frac{1}a + \frac{1}b + \frac{1}c = \frac{4}5$

Question

Solve in positive integers $\frac{1}a + \frac{1}b + \frac{1}c = \frac{4}5$ (i.e. find all triples $(a,b,c)$ of positive integers satisfying the equation).

The expression is equivalent to $5(bc + ac + ab) = 4abc.$ Suppose WLOG that $a=5k$ (at least one of $a,b,c$ must be divisible by $5$ and the equation is symmetric). After this, we'll just permute the elements to get all solutions. So we get $bc + 5kc + 5kb = 4kbc\Rightarrow (4k-1)(b+c) + (k+1)(b+c)=(4k-1)bc\Rightarrow (4k-1)(b-1)(c-1) = 4k-1 + (k+1)(b+c).$ However, this expression doesn't seem to factor nicely.

Answer 1

One way to do this is casework on the smallest denominator. Assume without loss of generality that $a\leq b\leq c$. Then $$\frac 45=\frac 1a+\frac1b+\frac1c\leq \frac 3a\implies a\leq \frac{15}4,$$ so $a\in\{1,2,3\}$. Clearly $1/1>4/5$, so $a\neq 1$.

If $a=2$, then $$\frac{b+c}{bc}=\frac 1b+\frac 1c=\frac45-\frac12=\frac3{10}.$$ This gives $3bc-10b-10c=0$, or $(3b-10)(3c-10)=100$. The only ways to factor $100$ as the product of two numbers which are $\equiv 2\pmod 3$ are $2\cdot 50$ and $5\cdot 20$, giving $(a,b,c)=(2,4,20)$ and $(a,b,c)=(2,5,10)$.

If $a=3$, then $$\frac{b+c}{bc}=\frac1b+\frac1c=\frac45-\frac13=\frac7{15}.$$ This gives $7bc-15b-15c=0$, or $(7b-15)(7c-15)=225$. There are no ways to factor $225$ as the product of two numbers which are $\equiv 6\pmod 7$ (none of the divisors $\leq 15$ of $225$ are $6\bmod 7$), so there are no solutions here.

In sum, the only solutions are $(2,4,20)$ and $(2,5,10)$.

Answer 2

The crucial observation is: if $a$ is too big, then $\frac{1}{a}$ is too small (and analogously for $b$ and $c$). Hence, the potential solutions involve natural numbers that have upper bounds.

It's not the most elegant solution, but since there are only three variables, you can do a case bash. WLOG, let's assume that $0 < a \leq b \leq c$, and we obtain all solutions by permutations of any solution $(a, b, c)$.

Notice that $a > 1$. So, consider $a = 2$. Then $$ \frac{1}{b} + \frac{1}{c} = \frac{3}{10}. $$ Since $\frac{1}{b} \leq \frac{3}{10}$, we must have $b \geq \frac{10}{3} > 3$. Now, since we've assumed that $b \leq c$, we have $\frac{1}{b} \geq \frac{1}{c}$, so $$ \frac{2}{b} \geq \frac{3}{10}, $$ hence $3b \leq 20$ or $b \leq \lfloor \frac{20}{3} \rfloor = 6$. Putting these bounds together, $b \in \{4, 5, 6\}$. With $b = 4$, we have $c = 20$, so $(a, b, c) = (2, 4, 20)$ is a solution. With $b = 5$, we have $c = 10$, so $(a, b, c) = (2, 5, 10)$ is another solution. With $b = 6$, $c \notin \mathbb{N}$, so we've found all solutions for $a=2$.

Now, consider $a = 3$. Then, $$ \frac{1}{b} + \frac{1}{c} = \frac{7}{15}. $$ Similar arguments yield $7b \leq 30$, or $b \in \{3, 4\}$. Unfortunately, neither of these possibilities yields a solution with $c \in \mathbb{N}$.

Finally, consider $a \geq 4$, so $$ \frac{1}{b} + \frac{1}{c} \leq \frac{11}{20}. $$ We have $b \geq 2$ and $11b \leq 40$ or $b \leq \frac{40}{11} < 4$, but then none of $a$, $b$, or $c$ is a multiple of $5$, so we can't have any more solutions.

Hence, there are $3! \cdot 2 = 12$ solutions, consisting of the $3! = 6$ permutations of each of $(2, 4, 20)$ and $(2, 5, 10)$.