This solution came from Sue Liu of Madras College, St Andrew's, Scotland.

The limerick gives three equations

${\displaystyle \begin{array}{cccc}\multicolumn{1}{c}{x+y+z}& =& 4& \hfill (1)\\ \multicolumn{1}{c}{\mathrm{xyz}}& =& 2& \hfill (2)\\ \multicolumn{1}{c}{x^2+y^2+z^2}& =& 6& \hfill (3)\end{array}}$

From (1)

${\displaystyle \begin{array}{cccc}\multicolumn{1}{c}{x+y}& =& 4-z& \hfill (4)\end{array}}$

and from (2)

${\displaystyle \begin{array}{cccc}\multicolumn{1}{c}{\mathrm{xy}}& =& \frac{2}{z}& \hfill (5)\end{array}}$

Using the identity $(x+y){}^2=x^2+y^2-2\mathrm{xy}$ equation (3) can be written as

${\displaystyle (x+y){}^2-2\mathrm{xy}+z^2=6}$

Hence

${\displaystyle (4-z){}^2-\frac{4}{z}+z^2=6.}$

Simplifying this equation gives

${\displaystyle z^3-4z^2+5z-2=0}$

${\displaystyle (z-1){}^2(z-2)=0}$

which gives the solutions $z=1$ and $z=2$.

Using equations (4) and (5) to find $x$ and $y$, when $z=1$ we get $x=1$ and $y=2$ or $x=2$ and $y=1$ and when $z=2$ we get $x=1$ and $y=1$.

The solutions are $(x,y,z)=(1,1,2)$ and $(1,2,1)$ and $(2,1,1).$

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