Find Positive Ints a,b,c Satisfying 1/ab+1/bc+1/ca=1/3

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The forum discussion focuses on finding all positive integers \(a, b, c\) that satisfy the equation \(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{1}{3}\) under the condition \(c \ge b \ge a\). Participants express enthusiasm for the solutions provided, indicating a collaborative effort in problem-solving. The discussion highlights the importance of logical reasoning and mathematical manipulation in deriving the integer solutions.

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Find all positive integers $a,\,b,\,c$ where $c \ge b \ge a$ and $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{1}{3}$.
 
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anemone said:
Find all positive integers $a,\,b,\,c$ where $c \ge b \ge a$ and $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{1}{3}$.

Rewrite the equation as $abc = 3(a+b+c) \leq 9c$. Therefore, $ab \leq 9$ (since $a,b,c > 0$).
Therefore, the possible sets for $(a,b)$ are $(1,u)$ where $u \in 1..9$, or
$(2,v)$ where $v \in 2..4$, or $(3,3)$. If we solve for $c$ in the equation $abc=3a+3b+3c$,
we have $c = \frac{3(a+b)}{ab-3}$. We can then eliminate base on $c$ is an integer, and $c \geq b > 0$,
so the possible values are:
\[
(1,4,15),(1,5,9),(1,6,7),(2,2,12),(2,3,5),(3,3,3).
\]
 
magneto said:
Rewrite the equation as $abc = 3(a+b+c) \leq 9c$. Therefore, $ab \leq 9$ (since $a,b,c > 0$).
Therefore, the possible sets for $(a,b)$ are $(1,u)$ where $u \in 1..9$, or
$(2,v)$ where $v \in 2..4$, or $(3,3)$. If we solve for $c$ in the equation $abc=3a+3b+3c$,
we have $c = \frac{3(a+b)}{ab-3}$. We can then eliminate base on $c$ is an integer, and $c \geq b > 0$,
so the possible values are:
\[
(1,4,15),(1,5,9),(1,6,7),(2,2,12),(2,3,5),(3,3,3).
\]

Awesome, awesome!(Party) Thanks for your neat solution!

Hey magneto, I get the feeling that you are striving to become another shining star at MHB!
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(Wink)
 

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