If z = 2, then the second equation gives (x + y)^{2} = 20. Since 20 is not a perfect square, no integer solutions are possible.

If x = y, then the second equation gives x^{2}(z + 2) = 20. Since the only perfect squares that divide 20 are 1 and 4, x = 1, -1, 2, and -2 with values for z of 18, 18, 3, and 3 respectively.

Therefore, the integer solutions for (x,y,z) are (1,1,18), (-1,-1,18), (2,2,3), and (-2,-2,3).

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