Page 2 of 2 FirstFirst 12
Results 16 to 19 of 19
Like Tree11Thanks

Math Help - show that a^2+b^2=3c^2 has no solutions in the integers

  1. #16
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: show that a^2+b^2=3c^2 has no solutions in the integers

    a^2 + b^2 = 3c^2 has no integral solutions can be proved with the algebra of odd and even:

    o+o=e, e+e=e, e+o=o
    oxo=o, exe=e, exo=e

    Substituting (a,b,c) odd or even into original equation gives:

    1) (o,o,o) → e=o, (o^2 +o^2 = 3o^2)
    2) (o,e,e) → o=e
    3) (e,e,0) → e=o
    4) (e,e,e) → e=e
    5) (o,o,e) → e=e
    6) (o,e,o) → o=o

    1), 2) and 3) ruled out because e=o.
    4) ruled out because dividing by 2^2 (again if nec) gives one of the other cases.

    5) let a=2l+1, b=2m+1, c=2n. Then,
    2l^2 + 2l + 2m^2 + 2m = 6n^2 1 → e=o

    6) let a=2l+1, b=2m+1, c=2n +1. Then,
    4l^2 + 4l + 4m^2+4m = 12n^2 +12n +1 → e=o

    Since that exhausts the possibilities for an integral solution, there isnt any.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: show that a^2+b^2=3c^2 has no solutions in the integers

    Quote Originally Posted by Hartlw View Post
    definition: a congruent b mod m iff m|(b-a) or a-b=mc for some c.

    If a,b, c are integers, the first step of link in post 6 says:

    a^2 + b^2 congruent 3c^2 mod 4, or a^2 + b^2 3c^2 = 4d for some d.

    Where does that come from, ie, how do you know a^2 + b^2 3c^2 is divisible by 4?
    Post 6 link starts off with xcongy mod 4 which is actually true for any integer if x=y. But 4 is significant for the case x= a^2 + b^2 and y = 3c^2 because it leads to a contradiction using arithmetic of congruence, and equivalence classes, I assume.
    Last edited by Hartlw; February 23rd 2013 at 11:29 AM. Reason: add "I assume"
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: show that a^2+b^2=3c^2 has no solutions in the integers

    Im going to stretch my welcome by adding these 2 short videos which give, in a simple manner, the modular arithmetic approach to solving this type of problem.
    I only meant to show the links. I don't know why big pictures of the site showed up. Don't mean to hog that much space. Try again:





    There, that's better. Wonder why the site showed up. Original link pasted underlined for some reason. Interesting.
    Damn, did it again. When I previewed post the links came back underlined and the sites showed up. Don't understand. Oh well.
    Last edited by Hartlw; February 23rd 2013 at 12:45 PM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: show that a^2+b^2=3c^2 has no solutions in the integers

    jojo7777777, you asked about Z (congruence arithmetic).

    A reasonable explanation is the first few pages of:

    www.math.ou.edu/~kmartin/nti/chap3.pdf

    (Post 16 is still a valid elementary algebra sol to OP if you don't have to do it with congruences.)
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: April 8th 2012, 08:05 PM
  2. Replies: 1
    Last Post: September 7th 2010, 08:30 PM
  3. Take integers a,b,c,d. Show that...
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 29th 2010, 10:41 AM
  4. Finding solutions in integers with constraints...
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 21st 2009, 10:41 PM
  5. Show that for all integers r,s
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 15th 2008, 02:35 PM

Search Tags


/mathhelpforum @mathhelpforum