(Number theory) Sum of three squares solution proof

  • Thread starter Xizel
  • Start date
  • #1
4
0

Homework Statement



Find all integer solutions to x2 + y2 + z2 = 51. Use "without loss of generality."

Homework Equations



The Attempt at a Solution



My informal proof attempt:

Let x, y, z be some integers such that x, y, z = (0 or 1 or 2 or 3) mod 4
Then x2, y2, y2 = (0 or 1) mod 4
So x2 + y2 + z2 = [ (0 or 1) + (0 or 1) + (0 or 1) ] mod 4
Since 51 = 3 (mod 4) = x2 + y2 + z2, then x, y, z = (1 or 3) mod 4

It is obvious that the solution is the permutations of ##\pm1, \pm1, \pm7##.

It's my first proof course and I'm a little shaky. Is my logic correct? I feel like I took a leap from "Since 51..." to the solution, is there a more formal way to write that? I'm also not sure how to use wlog here. Thanks.
 

Answers and Replies

  • #3
4
0
I have a additional solution.

If my logic is correct, then x, y, z can take on values of 1, 3, 5, 7. I gave it some thought and I'm not seeing it :nb)
 
  • #4
15,095
12,768
##2\cdot 25 + 1##
The remark "use w.l.o.g." probably refers to the assumption ##x \geq y \geq z \geq 0## which you could make.
 

Related Threads on (Number theory) Sum of three squares solution proof

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
4K
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
Top