Prove 2/5 perfect squares must be even to have their sum equal odd

Click For Summary

Homework Help Overview

The discussion revolves around proving that in the equation a² + b² + c² + d² + e² = f², at least two of the six variables must be even for the sum to be odd. The subject area includes number theory and properties of quadratic residues.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to prove the statement by considering cases based on the parity of f. They explore the implications of f being even or odd and question the validity of their approach. Some participants suggest using modular arithmetic, specifically modulo 8, to derive a contradiction.

Discussion Status

Participants are actively engaging with the problem, offering hints and discussing the utility of different modular bases. There is an acknowledgment of the original poster's method as workable, and suggestions have been made to refine their approach through modular considerations.

Contextual Notes

The original poster expresses uncertainty about their method and seeks guidance, indicating a need for clarification on how to effectively use modular arithmetic in this context.

cap.r
Messages
64
Reaction score
0

Homework Statement


given the equality a2+b2+c2+d2+e2=f2

prove 2 out of the the 6 variables must be even.

Homework Equations


can use quadratic residues and primitive roots if it helps but don't think i need them.

The Attempt at a Solution


assume f is even. then f2 is even. and not all 5 numbers on the left can be odd or else we would have odd=even. so at least one even on the LHS completes this case.

assume f is odd. then f2 is odd. so the LHS must have a odd number of odd numbers. 1,3,5 of these numbers must be odd. if it's 1,3 then we have at least 2 even and are done.

so now i need to prove by contradiction that not all 5 can be odd. this is where i am stuck and am thinking maybe the whole method is wrong.

a hint would be nice, thanks.
 
Last edited:
Physics news on Phys.org
Your method is perfectly workable. To obtain your contradiction in the last case try to consider the equality modulo 8 (HINT: If a is odd, then you know a^2 mod 8).
 
ok so RHS is 1 mod 8 and LHS is 5 mod 8. easy, but how did you know to use mod 8? i have noticed that mod 8,6,4 are particularly important. but i never know which one to use. i have actually considered mod 4 here but that didn't give me any results.
 
Mostly experience I guess. I knew that mod 8 have very few quadratic residues, and that 1 is the only odd one so it seemed like a natural choice. I also considered mod 4 briefly first, but saw that due to the fact that the left hand side wrapped around too early no contradiction could be obtained, but as mod 8 has very similar behavior for odd numbers I just thought of it. There are some quadratic residues that are often helpful. For instance working mod 2^n often gives good results, especially when dealing with powers of numbers whose parity you know.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
23
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
9K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
2K