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Let $a$, $b$, $c$ be three positive integer numbers, with $a \ge b$, such that $\frac{1}{a}+\frac{1}{b}+\frac{2}{c}$ is integer.

I've written a program to find the solutions for $a, b, c \le 1000$, and I've got:

a b c
1 1 1
1 1 2
2 1 4
2 2 1
2 2 2
3 1 3
3 2 12
3 3 6
4 2 8
4 4 4
6 2 6
6 3 4
6 6 3
10 2 5
12 4 3

But I don't see any proof for the fact that those are the only solutions. Can you help me? Thanks!

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  • $\begingroup$ You've performed a brute-force proof that no other integer solutions with a,b,c<=1000 exist, barring a programming error. $\endgroup$ – Ed Pegg Jun 3 '18 at 17:10
  • $\begingroup$ What about trying it for some more values, maybe we can be specific that there are no more solutions. $\endgroup$ – Love Invariants Jun 3 '18 at 17:13
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    $\begingroup$ If they're too large, it'll be strictly between $0$ and $1$ and thus not an integer $\endgroup$ – Akiva Weinberger Jun 3 '18 at 18:05
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Put $n = \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}= \dfrac{2ab+bc+ca}{abc}\implies 2ab+bc+ca = nabc\implies c(a+b-nab)= -2ab\implies c = \dfrac{2ab}{nab-a-b}$. We have: $nab-a-b \ge nab-2ab = (n-2)ab\implies c \le \dfrac{2ab}{(n-2)ab}= \dfrac{2}{n-2}\le 1$ for $n \ge 4$. This means $n = 1,2, 3$. Can you take it from here ? It's almost done.

Continuation of work after initial post:

If $n = 1 \implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c} = 1$. Since $a, b,c \ge 1$, none of them can be $1$, and $c \ge 3$. Thus if $c = 3$, then $\dfrac{1}{a}+\dfrac{1}{b} = \dfrac{1}{3}\implies \dfrac{1}{3} \le \dfrac{2}{b}\implies b \le 6\implies b = 2,3,4,5,6$ . By plug in each value of $b = 2,3,4,5,6$, only $b = 6$ yields $a = 6$ and is an integer. Thus $(n,a,b,c) = (1,6,6,3)$ is a solution. I think you can continue this work to obtain all the solutions of the equation above.

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  • $\begingroup$ It's great that you have found a restriction for $n$, but I don't think it's enough to find $a$, $b$, $c$. Can you explain me? $\endgroup$ – Iulian Oleniuc Jun 3 '18 at 17:46
  • $\begingroup$ “This means $n=1,2,3$.” —For $a=b=c=1$, $n=4$. $\endgroup$ – celtschk Jun 4 '18 at 5:31
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If $c > 12$ then $0< \frac 2c < \frac 2{12}$ so $1 > \frac 1a + \frac 1b > \frac {5}{6}$.

wolog $b \ge a$ so $\frac 1a \ge \frac {5}{12}$ while $\frac 1b \le \frac 5{12}$. So $a \le \frac {5}{12}$. So $a \le 2$ so $a= 2$ or $1$.

If $a=2$ then $\frac 12 + \frac 1b >\frac {5}{6}$ and $\frac 1b > \frac {1}{3}$ the $b < 3$ so $b \le 2$. So $2 \ge b \ge a = 2$. so $b = a = 2$.

But $\frac 12 + \frac 12 = 1 \ge 1$. So that's impossible.

If $a = 1$ then $\frac 1b + \frac 2c$ is an integer so $1 > \frac 1b > \frac {c-2}{c} > \frac 12$ which is not possible.

So $c \le 12$ which give you the terms you found.

We can continue in that manner.

If $c = 12$ then $\frac 1a + \frac 1b = \frac 56$ (or $1 \frac 56$)

Wolog $b > a$ so $\frac 1a \ge \frac 5{12}$ and $\frac 1b \le \frac 5{12}$ so $a \le 2$. If $a = 2$ then $\frac 12 + \frac 1b = \frac 56$ so $\frac 1b = \frac 13$.

If $a = 1$ then $\frac 1b = k + \frac 56$ must be a integer which is impossible.

And so on.....

If $c = 11$ then $\frac 1a + \frac 1b = \frac {9}{11}$ and so

wolog $b > a$ then $\frac 1a \le {10}{11}$ so $a = 2$ or $a = 1$ and if $a = 1$ we must have $\frab 1b + \frac 2{11}$ is an iteger which is impossible.

If $a = 2$ then $\frac 1b + \frac 12 = \frac {9}{11}$ so $\frac 1b = \frac 7{22}$ which is not possible.

....

I'll skip $c = 10,9$ but

If $c= 8$ then $\frac 1a + \frac 1b = \frac 68 =\frac 34$ and

wolog $b \ge a$ so $\frac 1a \ge 38$ and $a \le \frac 83$ so $a= 2$ or $a=1$. Again $a = 1$ will give us an impossible $\frac 1b + \frac 28$ is an integer.

But $a = 2$ means $\frac 12 + \frac 1b = \frac 34$ and $b = 4$.

And so on....

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