Question

3 positive integers multiply together to give S, six times their sum.
One of them is the sum of the other two.
Possible values of S?

Answer 1

Let the three integers be \(x, y, x+y\). Then we have the relation\[xy(x+y)=6(x+y+x+y)=12x+12y \]So\[x^2y+xy^2=12x+12y\]From here we can get a quadratic equation.\[yx^2+(y^2-12)x-12y\]Using the quadratic formula, we find the roots are given by\[\large {-(y^2-12)\pm\sqrt{(y^2-12)^2-4(-12y^2)}\over 2y}\]Simplifying this, we get\[\large {12-y^2 \pm \sqrt{y^4-24y^2+144+48y^2}\over2y}\]\[\large {12-y^2 \pm \sqrt{y^4+24y^2+144}\over2y}\]\[\large {12-y^2 \pm \sqrt{(y^2+12)^2}\over2y}\]Thus, our solutions are \[x={12-y^2+y^2+12 \over 2y},\qquad {12-y^2 -y^2-12 \over 2y}\]Or rather, \[x={24 \over 2y}={12 \over y},\qquad x=-{2y^2 \over 2y^2}={-1}\]However, \(x>0\) so we can discard the solution on the right.

Hence, all positive integral solutions are \((1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1)\)

Answer 2

Alternatively, we could have noticed that \[xy(x+y)-12x-12y=xy(x+y)-12(x+y)\]So it can factor as \[(x+y)(xy-12)=0\]And then solve from there.