If x=2, LHS is even so z=2 but 74y=1660 has no integral solutions.
That makes x odd and mod4 y+1=z^3 or y-1=z^3. In either case y and z are opposite parity.
Say z=2 and x(x+37y)=1664=128*13. x=13 will serve but (13+37y)=128 allows no solutions. That makes y=2.
Substitute that value, consider the equation as quadratic in x and look at the discriminant which must be a square.
We have 37^2 + z^3 + 1656 = k^2 which simplified and rearranged reads z^3 = k^2 - 55^2 or z^3 = (k+55)(k-55). Given prime z this means z^2 = (k+55) and z = (k-55). Subtracting gives z(z-1)=110 and z=11.
Plug in z=11 and solve the quadratic giving x=29 as the only positive root.
Checking, 29*29 + 37*29*2 = 841 + 2146 = 2987 and 11*11*11 + 1656 = 1331 + 1656 = 2987.
(x,y,z) = (29,2,11)
Posted by xdog
on 2014-03-12 16:58:59