All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
OOO3= one out of three (Posted on 2010-11-04) Difficulty: 4 of 5
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

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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


finds
x              y             z
1             28            35
4             25            37
5             7             44
5             31            32
7             19            40
11            17            40
16            23            35
19            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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information