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

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    given the equality a2+b2+c2+d2+e2=f2

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


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


    3. 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: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2
    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).
     
  4. Nov 9, 2009 #3
    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.
     
  5. Nov 9, 2009 #4
    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.
     
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