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Find all positive integer solution to $abc-2=a+b+c$.

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None of the variables can be greater than $4$, and at least two of them have to be greater than $1$. Also, by looking at the equation modulo 2, you can not have exactly one of them being odd. Finally, this equation is symmetric in its variables, which means I just have to say which three numbers I pick, not which one I assign to which variable. A brute force tactic doesn't take too long from here.

First we go with all the variables being odd. There are only two cases conforming to the list of demands in the last paragraph, and that is $(3, 3, 1)$ and $(3, 3, 3)$. The first set does solve the equation, the second one doesn't.

Next is two of them being odd. Setting the even number to $4$ makes the left side too large, so we need the even number to be $2$. The only set left is then $(2, 3, 1)$, which does not solve the equation.

Lastly, the case of all the digits even. There are four of these sets, depending on how many $2$s and $4$s we use. The sets $(4, 4, 4)$, $(4, 4, 2)$ and $(4, 2, 2)$ don't lead to a solution, but the set $(2, 2, 2)$ does.

All in all we have 2 sets of solution values, which can be distributed over the variables in a total of four different ways.

Edit It has been made known to me, thanks to Ross Pure, that I was a bit hasty to exclude the case of one variable being equal to $5$. So there is a set $(5, 2, 1)$, which generates another 6 solutions, depending on which number you give to which variable.

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I can't comment but in response to Arthur's answer: it's not so that the variables can't be greater than 4; (5,2,1) works.

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You're right, I somehow calculated that to yield $8 = 7$ and discarded it without a second thought. –  Arthur Feb 18 '13 at 12:37
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