# find a,b,c

#### Albert

##### Well-known member
$a,b,c \in N$

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

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

$find :a,b,c$

#### mathworker

##### Active member
Since modulo is "zero" there is no remainder.
$$\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}$$ is not a fraction
$$\displaystyle abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$$ is not a fraction
So the little terms must sum it up to zero or 1 so,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$$(let)
$$\displaystyle \frac{ab+bc+ca-1}{abc}=0$$
$$\displaystyle ab+bc+ca=1$$ which certainly can't be true
If,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1$$
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
But as $$\displaystyle a,b,c>1$$ $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ 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?

#### mathbalarka

##### Well-known member
MHB Math Helper
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!

#### mathworker

##### Active member
yeah I got it
.....has a maxmum value of 1.....
max value is not 1 $$\displaystyle its \frac{13}{12}$$

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But,I guess (2,3,5) is the only solution....

#### Albert

##### Well-known member
But,I guess (2,3,5) is the only solution....
yes (2,3,5) is the only solution..,but how to get the answer ?

#### mathworker

##### Active member
In,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$\displaystyle c=5$$

#### Albert

##### Well-known member
In,
$$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$\displaystyle c=5$$
very nice solution