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$a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc$
How many triples $(a,b,c)$ satisfies the statement? Here $a,b,c > 1$.

It is easy to simplify the statement to
$$ab + bc + ca = 2 + abc.$$ But now how to proceed I don't know. I think this is somehow related to stars and bars theorem but I don't know how to convert this to that problem. Any hint will be helpful.

Source this is a problem from BdMO 2015 Dhaka regional.

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    Try to divide by $abc$ and then use inequalities that one of a,b,c should be $\leq$3 – 2017-01-21

2 Answers 2

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If $a$, $b$, and $c$ are required to only be positive integers and some of them is $1$, then we have a unique solution $(a,b,c)=(1,1,1)$. For solutions with $a,b,c>1$, note that $$(a-1)(b-1)(c-1)=abc-bc-ca-ab+a+b+c-1=a+b+c-3\,.$$ Set $x:=a-1$, $y:=b-1$, and $z:=c-1$. Therefore, $$xyz=x+y+z\,.$$ Without loss of generality, suppose that $x\leq y\leq z$ (noting that they are positive integers). Now, we have $$y+z=x(yz-1)\geq yz-1\,,\text{or }(y-1)(z-1)\leq 2\,.$$ If $(y-1)(z-1)=0$, then $y=1$, making $x=1$ as well, so $$z=xyz=x+y+z=2+z\,$$ which is absurd. If $(y-1)(z-1)=1$, then $y=2$ and $z=2$, so $$4x=xyz=x+y+z=x+4\,,$$ leading to $x\notin\mathbb{Z}$, which is a contradiction. Thus, $(y-1)(z-1)=2$, and so $y=2$ and $z=3$. Ergo, $$6x=xyz=x+y+z=x+5\,,\text{ or }x=1\,.$$ Consequently, $(a,b,c)=(2,3,4)$ is the only solution, up to permutation.

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    This seems harder to me ... I think there might be some easier solution... – 2017-01-22
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    If you use arberavdullahu's hint carefully, it leads to a simpler solution. (In fact, it can be shown using the hint that one of $a$, $b$, and $c$ must be $2$.) I am merely providing a different route. You should try harder with your own question, especially if you think there is a simple solution. – 2017-01-22
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I'm also a Bangladeshi, reading in class IX in DRMC. Probably the problem was:

How many solution triads $(a,b,c)$ are there for the equation $a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc?$ Where $(a,b,c)$ are positive integers and $a,b,c>1$.

After trying a lot, I've got a simple solution. Suppose that, $2≤a≤b≤c$.

Note that,

$$ab+bc+ca-2<3bc$$

i.e. $$abc<3bc$$ So, $a=2$

Set the value of $a$. Therefore,

$$2b+2c = 2+bc$$

Now, $2b+2c=2+bc>bc$

Hence, $(b-2)(c-2)<4$

Solve the equation. Finally, $(a,b,c)=(2,3,4)$ is the only solution.