MHB What are the values of a, b, and c in the following equations?

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The discussion revolves around finding natural numbers a, b, and c that satisfy the equations 1<a<b<c and (ab-1)(bc-1)(ca-1) mod (abc)=0. Initial attempts to derive a solution led to the conclusion that no such values exist, as the derived equations suggested contradictions. However, through further exploration, it was determined that (2, 3, 5) is indeed a valid solution. The reasoning involved substituting values into the equations and solving for c, confirming that the left-hand side must exceed one when a and b are set to 2 and 3. Ultimately, the consensus is that (2, 3, 5) is the only solution that meets the criteria.
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$a,b,c \in N$

(1) $1<a<b<c$

(2)$(ab-1)(bc-1)(ca-1) \,\, mod \,\, (abc)=0$

$find :a,b,c$
 
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Since modulo is "zero" there is no remainder.
$$\frac{(ab-1)(bc-1)(ca-1)}{abc}$$ is not a fraction
$$abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$$ is not a fraction
So the little terms must sum it up to zero or 1 so,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$$(let)
$$\frac{ab+bc+ca-1}{abc}=0$$
$$ab+bc+ca=1$$ which certainly can't be true
If,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
But as $$ a,b,c>1$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?:confused:
 
I am just going to leave this one here : (a, b, c) = (2, 3, 5). (*) Can you see where you went wrong now?

(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!
 
yeah I got it
...has a maxmum value of 1...
max value is not 1 $$its \frac{13}{12}$$

- - - Updated - - -

But,I guess (2,3,5) is the only solution...(Smirk)
 
mathworker said:
But,I guess (2,3,5) is the only solution...(Smirk)
yes (2,3,5) is the only solution..,but how to get the answer ?
 
In,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$c=5$$
 
mathworker said:
In,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$c=5$$
very nice solution(Yes)
 

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