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
(y-2009)^2 + (z-2009)^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.
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Posted by Jer
on 2015-06-18 12:33:08 |
Solution
