presidentabrahamlincoln

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Let w, x, y, and z belong to the complete set of integers.
If each of wxy + z^2, wxz + y^2, wyz + x^2, xyz + w^2 is divisible by 4, show that
w^3 + x^3 + y^3 + z^3 is divisible by 4.

9(9  9)! = 9

Please ignore the prior post.
Let the remaining matchsticks be located at these coordinates on the xyaxes:
(0, 0) to (0, 1)
(0, 1) to (0, 2)
(0, 2) to (0, 3)
(2, 0) to (2, 1)
(2, 1) to (2, 2)
(2, 2) to (2, 3)
(3, 1) to (3, 2)
(3, 2) to (3, 3)
(0, 0) to (1, 0)
(1, 0) to (2, 0)
(0, 1) to (1, 1)
(1, 1) to (2, 1)
(2, 1) to (3, 1)
(0, 2) to (1, 2)
(1, 2) to (2, 2)
(2, 2) to (3, 2)
(0, 3) to (1, 3)
(1, 3) to (2, 3)

On 12/1/2015 at 8:13 AM, bonanova said:Any other solutions?
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Show 4 divides this algebraic expression.
in New Logic/Math Puzzles
Posted · Edited by presidentabrahamlincoln
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Let w, x, y, and z belong to the complete set of integers.
If each of wxy + z^2, wxz + y^2, wyz + x^2, xyz + w^2 is divisible by 4, show that
w^3 + x^3 + y^3 + z^3 is divisible by 4.