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 OOO3= one out of three (Posted on 2010-11-04)
Given an equation x^2+y^2+z^2=2010.
1. Prove that in every triplet of integers satisfying the above equation one number has to be even and two others odd.
2. How many integer solutions are there ?

Warning: 1 is very easy, 2 is quite tricky.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 solutions | Comment 2 of 8 |

1. If all x, y and z were even the sum would be a multiple of 4, which 2010 is not. If one were odd and the others even, or all were odd, the sum would be odd, which 2010 is not. So two are odd and one is even.

2.

CLS
DEFDBL A-Z
FOR x = 0 TO 45
x2 = x * x
FOR y = x TO 45
y2 = y * y
FOR z = y TO 45
z2 = z * z

IF x2 + y2 + z2 = 2010 THEN PRINT x, y, z

NEXT
NEXT
NEXT

`findsx              y             z 1             28            354             25            375             7             445             31            327             19            4011            17            4016            23            3519            25            32`

These are the 8 non-trivial variations.

Each of these can be permuted 6 ways, producing 48 positive integral solutions.

Each of x, y and z can, independently, be made positive or negative, multiplying the solutions by 2^3 = 8, making the final number of integer solutions 8*6*8 = 384.

 Posted by Charlie on 2010-11-04 18:44:03

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