Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

Substituting (m, n) = (x+y, x-y) in the given equation and simplifying, we obtain:

m^2(z+2) + n^2(z-2) = 80 .......(i)

m^2(z+2) - n^2(z-2) = 80 ......(ii)

Subtracting (ii) from (i), we obtain:

n^2(z-2) = 0, so that:

Either, n=0; Or, z=2

If z=2, then from (i), we obtain:

4*m^2 = 80, so that: m= 2V5, which is a contradiction- since, by the problem m must be an integer.

If n = 0, then x=y(=p, say); so that: m=2p

Thus, substituting (n, m) = (0, 2p) in (ii), we obtain:

4*p^2(z+2) = 80

or, p^2(z+2) = 20

or, (p, z) = (1, 18); (-1, 18); (2, 3); (-2, 3)

or, (x, y, z) = (1, 1, 18); (-1, -1, 18); (2, 2, 3); (-2, -2, 3)

Thus, (x, y, z) = (1, 1, 18); (-1, -1, 18); (2, 2, 3); (-2, -2, 3)

gives all possible solutions to the given problem.