Solve the second equation for x, and square it
x^2 = 4018^2 + y^2 + z^2  8036y  8036z + 2yz
substitute into the first:
4018^2 + 2y^2 + 2z^2  8036y  8036z + 2yz = 2yz + 2
divide by 2, then complete the squares
(y2009)^2 + (z2009)^2 = 1
This is a circle centered at (y,z)=(2009,2009) with r=1
There are obviously only 4 integer solutions, they each lead to a value of x. These solutions (x,y,z) are:
(1,2008,2009)
(1,2009,2008)
(1,2010,2009)
(1,2009,2010)
The symmetry of interchanging y and z was clear from the start, but turned out not to be needed.

Posted by Jer
on 20150618 12:33:08 