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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$a,b,c \in N$

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

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

$find :a,b,c$

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

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

$find :a,b,c$

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

$a,b,c \in N$

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

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

$find :a,b,c$

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

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

$find :a,b,c$

- May 31, 2013

- 118

\(\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}\) 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?

- Mar 22, 2013

- 573

(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!

- May 31, 2013

- 118

max value is not 1 \(\displaystyle its \frac{13}{12}\).....has a maxmum value of 1.....

- - - Updated - - -

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

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- #5

- Jan 25, 2013

- 1,225

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

- May 31, 2013

- 118

\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)

L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation

\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)

\(\displaystyle c=5\)

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- #7

- Jan 25, 2013

- 1,225

very nice solution

\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)

L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation

\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)

\(\displaystyle c=5\)