Results 1 to 2 of 2

Math Help - Pythagorean triple

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    16

    Pythagorean triple

    This problem has got me stumped.

    Let (x,y,z) be a primitive Pythagorean triple. Prove that one of the integers in the triple is divisible by 3, one is divisible by 4, and one is divisible by 5.

    3,4,5 do not have to divide different elements of the triple.

    I mean we know that by definition a primitive Pythagorean triple has to satisfy x^2+y^2=z^2. We also know that one leg is odd, the other is even, and that the hypotenuse is odd. Also, gcd(x,y,z)=1.

    I tried isolating one variable and then trying to show it is divisible by 3,4,5, but am running into trouble.

    Help would be awesome. Thanks.
    Last edited by mr fantastic; May 22nd 2009 at 02:54 PM. Reason: Restored original question deleted by OP
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by curiousmuch View Post
    This problem has got me stumped.

    Let (x,y,z) be a primitive Pythagorean triple. Prove that one of the integers in the triple is divisible by 3, one is divisible by 4, and one is divisible by 5.

    3,4,5 do not have to divide different elements of the triple.

    I mean we know that by definition a primitive Pythagorean triple has to satisfy x^2+y^2=z^2. We also know that one leg is odd, the other is even, and that the hypotenuse is odd. Also, gcd(x,y,z)=1.

    I tried isolating one variable and then trying to show it is divisible by 3,4,5, but am running into trouble.

    Help would be awesome. Thanks.
    Hi curiousmuch.

    Let

    x=m^2-n^2

    y=2mn

    z=m^2+n^2

    m and n cannot be both odd, otherwise x,y,z would all be even. Hence one of m and n must be even. Hence y is always divisible by 4.

    If one of m and n is divisible by 3, then y would also be divisible by 3. Suppose neither m nor n is a multiple of 3. Then 3 does not divide m^2 or n^2 either. Now, either m^2\equiv n^2\pmod3 or m^2\not\equiv n^2\pmod3. In the former case, x=m^2-n^2 is divisible by 3. In the latter case, we have m^2\equiv\pm1\pmod3 and n^2\equiv\mp1\pmod3. \therefore 3 divides m^2+n^2=z.

    If one of m and n is divisible by 5, then y would also be divisible by 5. Suppose 5 does not divide m or n. By Fermat’s little theorem, m^4\equiv1\pmod5 and n^4\equiv1\pmod5. Hence 5 divides m^4-n^4=(m^2-n^2)(m^2+n^2)=xz. As 5 is prime, this means 5 divides either x or z.
    Last edited by TheAbstractionist; May 19th 2009 at 02:32 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. About Pythagorean triple...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: August 8th 2010, 09:15 PM
  2. Pythagorean triple
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: May 18th 2009, 09:58 PM
  3. Pythagorean Triple
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: May 13th 2009, 10:44 AM
  4. Pythagorean triple in (a,a+1,c)
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: March 22nd 2009, 07:12 PM
  5. Pythagorean triple
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 8th 2009, 07:51 PM

Search Tags


/mathhelpforum @mathhelpforum