Integer solutions to fraction equation

Question

I am having trouble finding all solutions (or at least proving I have all the solutions already).

The equation is $$\frac{1}{a}+\frac{1}{b}+\frac{2}{c}=1$$

*a,b,c are positive I tried to base it on the solutions of a similar equation with the last term a 1 instead of a 2.

Answer 1

Let $a\geq b$.

Hence, $$1=\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\leq\frac{2}{b}+\frac{2}{c},$$ which gives $$bc\leq2b+2c$$ or $$(b-2)(c-2)\leq4.$$ Now, if $b\geq3$ we obtain: $$c-2\leq\frac{4}{b-2}\leq4,$$ which gives $c\leq6$ and the rest is smooth.

For example, the case $b=1$ is impossible, but for $b=2$ we obtain $$\frac{1}{2}=\frac{1}{a}+\frac{2}{c}$$ or $$ac=2c+4a$$ or $$ac-2c-4a+8=8$$ or $$(a-2)(c-4)=8$$ and solve some systems:

1. $a-2=1$ and $c-4=8$ gives $(3,2,12)$;

2. $a-2=2$ and $c-4=4$ gives $(4,2,8)$;

3. $a-2=4$ and $c-4=2$ gives $(6,2,6)$ and

4. $a=2=8$ and $c-4=1$ gives $(10,2,5).$

Now, for $a\geq b\geq3$ we need to check $c\in\{3,4,5,6\}$, which is for you.

For the full ending of the solution it's enough to add triples $(b,a,c)$.

Answer 2

Notice each $\frac 1a, \frac 1b, \frac 1c$ average $\frac 13$ but $\frac 1a,\frac 1b$ have a maximum value of $\frac 12$ and $\frac 2c$ has a maximum value of $\frac 23$.

So Case 1: Each term is equal to the average:

$(a,b,c) = 3,3,6$.

Case 2: $\frac 1a$ or $\frac 1b$ is greater than the average.

Then $a = 2$ or $b = 2$.(wolog $a=2$)

Then $a = 2$ and $\frac 1b + \frac 2c = \frac 12$.

$\frac 2c = \frac 12 - \frac 1b = \frac {b-2}{2b}$

$c = \frac {4b}{b-2}$. $\gcd(b-2, b) \le 2$ so $b-2|4b$ means $b-2$ is a power of $2$. $b-2$ is a higher power of $4$ than $2$, then $b= b-2 + 2$ is only divisible by $2$ and not $4$ so $b-2$ may equal $8,4,2$ or $1$.

So $a = 2$ yields $b = 10; c=5$ or $b=6, c= 6$ or $b=4, c = 8$ or $b=3, c= 12$.

That was the hardest case.

Case 3: $\frac 2c$ is greater than the average.

The $c = 3,4$ or $5$

For each of those you get $\frac 1a + \frac 1b = 1-\frac 1c$

You can find the average of $\frac 1a, \frac 1b$. And thus a maximum testing value. So test each $a= $ minimum value to $a =$ maximum value.

And that will be all possible solutions.