If x+y+z=19,xyz=144 then find the value of x,y,z
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I don't have any great insight to offer, though I wouldn't be surprised if there is one to be had. I can offer what I think is a systematic way to generate all possible solutions, assuming that x y and z are positive integers as stated. The first step is to factor 144 completely
$\displaystyle 144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$
Then generate the set of possible factors of 144. I did this "by the seat of my pants," but I think the list is complete.
$\displaystyle \left\{ {1,2,3,4,6,8,9,12,16,18,24,36,48,72,144} \right\}$
Obviously all numbers greater than 16 aren't useful for this problem. There are not 2 other integers you can add to 18, let alone 24, to get 19. That leaves
$\displaystyle \left\{ {1,2,3,4,6,8,9,12,16} \right\}$
Now suppose $\displaystyle x = 1$. I then write $\displaystyle yz = 144$ and
$\displaystyle \begin{array}{l} y + z = 18\\ z = 18 - y \end{array}$
From which I can get a quadratic equation
$\displaystyle \begin{array}{l} y\left( {18 - y} \right) = 144\\ {y^2} - 18y + 144 = 0 \end{array}$
That has no real roots. This means $\displaystyle x \ne 1$ and by symmetry neither does y or z; also, none of them x y or z can be 16 because then one of the others would have to be 1.
I haven't followed this procedure out to the bitter end, but I'm satisfied that the two solutions already given, namely $\displaystyle \left\{ {2,8,9} \right\}$ and $\displaystyle \left\{ {3,4,12} \right\}$ are the only solutions that exist.
BTW, it did occur to me to use combinations with repetition to find the number of integer solutions to x + y + z = 19, but I doubt that would've saved any time.