SOLUTION: if z=x+iy , z* conj(z)^3+conj(z)*z^3=350, then (x,y)=..[(2,1) ,(3,4) ,(4,3) ,(5,4)]

z*conj(z)^3+conj(z)*z^3 = z*conj(z)*(z^2 + (conj((z))^2) = |z|^2*(2x^2 - 2y^2).


So, our equation takes the form

      |z|^2*(2x^2 - 2y^2) = 350,  

or

     |z|^2*(x^2 - y^2) = 175.    (2)


In this problem,  |z|^2  and (x^2-y^2) are integer numbers;  so, our equation (2)  is an equation in integer numbers.


We have |z|^2 =  5  for  z = (2,1);

        |z|^2 = 25  for  z = (3,4);

        |z|^2 = 25  for  z = (4,3);

        |z|^2 = 41  for  z = (5,4).


Therefore, based on properties of divisibility, we see that (5,4) does not work.

Also,  x^2 - y^2 = 9 - 16  is negative number, so,  (3,4) does not work, too.

Next, for (2,1)  x^2 - y^2 = 2^2-1^2 = 4 - 1 = 3 does not divide 175, so (2,1) does not work.


The only remaining candidate is (4,3).  For it, |z|^2 = 25 and x^2 - y^2 = 16-9 = 7;  25*7 = 175.


So, the only pair which works is (4,3).


ANSWER.  Of four given pairs, the only solution to this problem is the pair (4,3).