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

The limerick gives three equations

x + y + z = 4 (1) xyz = 2 (2) x2 + y2 + z2 = 6 (3)

From (1)

x + y = 4 − z (4)

and from (2)

xy =  2 z (5)

Using the identity (x+y)²=x²+y²−2xy equation (3) can be written as

(x+y)2−2xy+z2=6

Hence

(4−z)2 −  4 z +z2 = 6.

Simplifying this equation gives

z3 − 4z2 + 5z −2 = 0

(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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