 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Linear and Quadratic Puzzle (Posted on 2015-06-18) Determine all possible integers x, y and z such that:
x2 + y2 + z2 = 2(yz + 1) and:
x + y + z = 4018

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 2
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.

 Posted by Jer on 2015-06-18 12:33:08 Please log in:

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