Results 1 to 6 of 6

Math Help - Number Theory , an easy proof .

  1. #1
    Super Member
    Joined
    Jan 2009
    Posts
    715

    Number Theory , an easy proof .

    Let's first recall some Pythagorean Triples :


     (3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61) etc .


    Can you see a pattern ? In every triple , we must find one integer from (x,y,z) , such that it is the mutiple of  5 .

     5,5,15,40,60 etc .


    I would like you to prove this fact , the proof shouldn't be so long , i believe ...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    Let (a,b,c) be a pythagorean triple. Observe that x^2 \equiv 0,1,4 (mod) 5.

    Suppose neither a,b,c\equiv 0 (mod) 5. Then a^2,b^2,c^2\equiv 1,4 (mod) 5

    But then we can never have a^2+b^2\equiv c^2 (mod) 5

    Contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by simplependulum View Post
    Let's first recall some Pythagorean Triples :


     (3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61) etc .


    Can you see a pattern ? In every triple , we must find one integer from (x,y,z) , such that it is the mutiple of  5 .

     5,5,15,40,60 etc .


    I would like you to prove this fact , the proof shouldn't be so long , i believe ...
    The integers m^2-n^2,\ 2mn,\ m^2+n^2 form a Pythagorean triple.

    Consider modulo 5.

    If m\equiv 0\ (mod\ 5) or n\equiv 0\ (mod\ 5), then 2mn\equiv 0\ (mod\ 5).

    If m\equiv\pm n\ (mod\ 5), then m^2-n^2\equiv 0\ (mod\ 5).

    If m\equiv\pm2\ (mod\ 5) and n\equiv\pm1\ (mod\ 5) or m\equiv\pm1\ (mod\ 5) and n\equiv\pm2\ (mod\ 5), then m^2+n^2\equiv 0\ (mod\ 5).

    This concludes all the cases.
    Last edited by alexmahone; August 2nd 2010 at 07:18 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Hello,
    The proof is trivial. Let x \equiv a \pmod{5}, y \equiv b \pmod{5}, z \equiv c \pmod{5}. Now consider a Pythagorean triple :

    x^2 + y^2 = z^2

    Since all terms a, b and c are modulo 5, we may equate and say that :

    a^2 + b^2 \equiv c^2 \pmod{5}

    According to the conjecture, this congruence may only be satisfied if and only if either a, b, c \equiv 0 \pmod{5} (if one of these integers is divisible by 5). Recall the quadratic residues modulo 5 :

    0^2 \equiv 0 \pmod{5}
    1^2 \equiv 1 \pmod{5}
    2^2 \equiv 4 \pmod{5}
    3^2 \equiv 4 \pmod{5}
    4^2 \equiv 1 \pmod{5}

    Now consider all the possibilities (I omitted those that didn't satisfy the conjecture because there's too much of them, this has been double-checked by computer) :

    0^2 + 0^2 = 0^2 \pmod{5}
    0^2 + 1^2 = 1^2 \pmod{5}
    0^2 + 1^2 = 4^2 \pmod{5}
    0^2 + 2^2 = 2^2 \pmod{5}
    0^2 + 2^2 = 3^2 \pmod{5}
    0^2 + 3^2 = 2^2 \pmod{5}
    0^2 + 3^2 = 3^2 \pmod{5}
    0^2 + 4^2 = 1^2 \pmod{5}
    0^2 + 4^2 = 4^2 \pmod{5}
    1^2 + 0^2 = 1^2 \pmod{5}
    1^2 + 0^2 = 4^2 \pmod{5}
    1^2 + 2^2 = 0^2 \pmod{5}
    1^2 + 3^2 = 0^2 \pmod{5}
    2^2 + 0^2 = 2^2 \pmod{5}
    2^2 + 0^2 = 3^2 \pmod{5}
    2^2 + 4^2 = 0^2 \pmod{5}
    3^2 + 0^2 = 2^2 \pmod{5}
    3^2 + 0^2 = 3^2 \pmod{5}
    3^2 + 1^2 = 0^2 \pmod{5}
    3^2 + 4^2 = 0^2 \pmod{5}
    4^2 + 0^2 = 1^2 \pmod{5}
    4^2 + 0^2 = 4^2 \pmod{5}
    4^2 + 2^2 = 0^2 \pmod{5}
    4^2 + 3^2 = 0^2 \pmod{5}
    2^2 + 1^2 = 0^2 \pmod{5}

    By analyzing the triples that do satisfy the conjecture, we find that all those triples have one, and only one, integer that is a multiple of 5 ... except the first one. However, apart from 0^2 + 0^2 = 0^2 which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by 25 cannot possibly be divisible by 25 (the proof is left as an exercise to the reader ).

    Therefore the conjecture is verified and valid.

    \mathfrak{QED}

    (okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2010
    From
    Israel
    Posts
    148
    Any integer is congruent to 0,\pm 1,\pm 2 modulo 5 (complete residue system). So any square is congruent modulo 5 to 0,\pm 1.
    Assume that both a^2 and b^2 are not congruent modulo 5 to 0, then a^2+b^2\equiv 0,2, or -2(mod\ 5).
    Since c^2=a^2+b^2 and c^2 is a square, we must have c^2\equiv 0(mod\ 5).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Thanks , indeed , all the solutions here are not so long . I want to include my proof in this post :


    Suppose all the primitive solutions of the equation  x^2 + y^2 = z^2 can be represented by  (m^2 - n^2 , 2mn , m^2 + n^2 ) .  x  is odd while  y is even ,  m , n are coprime and one odd one even . (It is true but the proof not included here . )

    Then take modulo 5 , we have

    m^2 - n^2 = (m-n)(m+n)

     m^2 + n^2 \equiv m^2 - 4n^2 = (m-2n)(m+2n)

    consider the product  xyz , assume  n is prime to  5 , otherwise ,  y is divisible by  5 .

    We have  xyz \equiv 2n(m-2n)(m-n)(m)(m+n)(m+2n) \equiv 2n^5 (s-2)(s-1)(s)(s+1)(s+2) \bmod{5} where  s \equiv m(n^{-1}) \bmod{5} . We can see that the product  (s-2)(s-1)(s)(s+1)(s+2) is divisible by  5 so is   xyz , this finishes the proof .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Easy character theory proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 4th 2010, 02:56 PM
  2. Number Theory Proof
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 12th 2008, 08:07 PM
  3. number theory proof
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: September 13th 2008, 06:17 PM
  4. Replies: 15
    Last Post: April 29th 2007, 03:42 PM
  5. Number theory proof
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: September 12th 2006, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum