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Prove a^2 + b^2 = 3(s^2 + t^2) implies both a and b must are divisible by 3

  1. Dec 20, 2007 #1
    I'm stuck on a seemingly simple part of a proof (a proof showing there are no non zero solutions of the equation a^2 + b^2 = 3(s^2 + t^2)

    at one step it says if

    a^2 + b^2 = 3(s^2 + t^2) this implies

    both a and b must be divisible by 3.
    I tried to prove this myself but have had no luck...

    any ideas?
     
  2. jcsd
  3. Dec 20, 2007 #2

    CRGreathouse

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    If a was divisible by 3 then a^2 would be divisible by 3 (and 9). If b wasn't, then b^2 wouldn't be and so of course a^2 + b^2 couldn't be. Now all you need to show is that if neither of the two are divisible by 3 then their sum can't be. Consider the equation mod 3 (or 9) and you should find the answer.
     
  4. Dec 20, 2007 #3
    I got as far as needing to show that if both a and b are not divisble by three then a^2 + b^2 must not be divisible by 3, but thats exactly where i got stuck,all attempts to show this have proved dead ends :(

    *edit*

    what i have managed to show is that if:

    a not divisible by 3 => a%3 = 1 or 2

    if a%3 = 1
    a^2 % 3 = 1

    if a%3 = 2
    a^2 % 3 = 1

    thus (a^2 + b^2) % 3 will always be 2 if a%3 and b%3 are non zero

    this seems sufficient...
     
    Last edited: Dec 20, 2007
  5. Dec 20, 2007 #4

    Gokul43201

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    Yes, that is sufficient.

    This leaves you with the final option: a and b are both divisible by 3.
     
  6. Dec 20, 2007 #5
    Ah okay, thanks for the pointer GreatHouse and thanks for the confirmation Gokul.

    :smile:
     
  7. Dec 21, 2007 #6
    a^2 == -b^2 mod 3

    if gcd(a,3)=1 ==> a^2 == 1 == -b^2 mod 3 ==> gcd(b,3)=1

    but b^2 == -1 mod 3 is a contradiction ==> a and b must be divisible by 3
     
    Last edited: Dec 21, 2007
  8. Dec 23, 2007 #7
    Ceres: a^2 + b^2 = 3(s^2 + t^2) this implies
    both a and b must be divisible by 3.

    Then if a and b are both divisible by 3, we now have 9(a'^2+b'^2) = 3(s^2+t^2), and this leads us to: 3(a'^2+b'^2) =s^2+t^2.

    This looks like the original problem reversed!(?)
     
    Last edited: Dec 23, 2007
  9. Dec 25, 2007 #8

    mathwonk

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    just work mod 3. i.e. the equation says you have a^2 +b^2 = 0 mod 3. trying a,b = 0,1,2, one sees nothing works except a=b=0.

    so in fact the c^2+d^2 part is a red herring.
     
  10. Dec 26, 2007 #9

    Gib Z

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    I can't tell if robert is just confused, or more likely, taking appreciation at the step of 'descent' =] .
     
  11. Dec 26, 2007 #10
    Giz Z: I can't tell if robert is just confused, or more likely, taking appreciation at the step of 'descent=].

    I was just hoping to go another step with the problem. INFINITE DESCENT? Yeah, just go look that up on Wikipedia, and by gosh! There is our problem.
     
  12. Dec 27, 2007 #11
    I think this is a simple proof, using Fermat's theorem.

     
  13. Dec 27, 2007 #12
    I was just trying to point out, as a hint, that the problem could be expanded to prove it had no non-trivial solutions. That is:

    a^2+b^2 not equal to 3(s^2+t^2) unless a=b=c=d=0.


    That was the intent. I guess it was not clear.
     
    Last edited: Dec 27, 2007
  14. Dec 28, 2007 #13

    Gib Z

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    When one uses a result much stronger then what is to be proven, It's sort of like using the mean value theorem to prove Roelle's theorem. Just an opinion though.
     
  15. Dec 28, 2007 #14
    I was just trying to help :cool:

     
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