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.

The squares of even numbers are even.  The squares of odd numbers are odd.  The sum of two odd numbers is even.  The sum of three odd numbers is therefore odd.  The sum of any number of even numbers is even.  2010 is even.  QED ? Anything more needed, e.g. that only even numbers have 2 as a factor (defintion of "even").

Is the "tricky" part of finding the answer to (2) that (a) the specs do not say if the same number may be REPEATED (i.e. whether or not x, y, and z must be different).  ALSO (b): may x, y, and z be NEGATIVE (their squares would of course all be positive, since we are limited to integers). 

Now to generate the sets.

If positive integers only, then there are eight triplets: 1,28,35  4,25,37  5,7,44  5,31,32  7,19,40  11,17,40  16,23,35  and 19,25,32.

If negative integers are allowed, then the triplets would be eight times as many, i.e. 64 triplets (each of the three values above in each triplet may be signed positive or negative -- for eight combinations where originally each counted once).

I am assuming that the SEQUENCE of each set is not counted more than once (these are clearly not different "integer solutions" (i.e. 11,17,40 and 17,40,11 are the same SET). If that is the "tricky" part, just multiply each sequence by 6 to count the permutations separately.

 

 

 

 

 

 

 

Edited on November 4, 2010, 5:52 pm

Posted by ed bottemiller on 2010-11-04 17:16:34