Results 1 to 3 of 3

Math Help - pythagorean triples

  1. #1
    Junior Member
    Joined
    Oct 2007
    From
    Nova Scotia
    Posts
    48

    pythagorean triples

    Show that the products of the lengths of the three sides is divisible by 60.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    60=2^2\cdot{3}\cdot{5}

    Suppose (x, y, z) is a pythagorean triple

    Then x^2+y^2=z^2

    Note that x^2\equiv{0;1}(\bmod. 3) , x^2\equiv{0;1}(\bmod. 4) and x^2\equiv{0;1;-1}(\bmod. 5) (these are the options)

    Suppose none of them is divisible by 3 then x^2\equiv{1}(\bmod{3}), y^2\equiv{1}(\bmod{3}) and z^2\equiv{1}(\bmod{3}) . But this is absurd since z^2=x^2+y^2\equiv{2}(\bmod{3})

    Thus at least one must be divisible by 3.

    Go over the other cases working in this way and you will get that at least one is divisible by 4 and 5, so the product is divisble by 60
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428

    ugerly proof

    I will prove that at least one of a, b and c must be a multiple of 4,5 and 3 and so their product must be a multiple of 60

    recall that a^2+b^2=c^2 =>

    c= r^2+s^2
    b=2rs
    a= r^2-s^2=(r+s)(r-s)

    for some natural r,s

    4.
    suppose that a, b and c have no common factor. Since b is even, either c or a must be odd. Since b^2 is also even, a^2 is odd iff c^2 is odd therefore both a and c are odd. Since c = r^2+s^2, one of r and s is odd and the other even. So 4|b since b=2rs.
    Since every triple is a multiple of a triple with no common factor, 4|b in general.

    3.
    if 3|r or 3|s then 3|b
    if r \equiv s mod 3 then 3|(r-s) => 3|a
    This leaves the possibilities r \equiv 2 mod 3 and s\equiv1 mod 3 or vica versa. Either way, 3|(r+s) => 3|a

    5.
    if 5|r or 5|s then 5|b
    if r \equiv s mod 5 then 5|(r-s) => 5|a
    <br />
r\equiv4 mod 5, s\equiv 3 mod 5: r = 5x+4, s = 5y+3. r^2+s^2
    = 25x^2+40x+16+25y^2+30y+9
    = 25x^2+40x+25y^2+30y+25\equiv0 mod 5, so 5|c
    The previous approach can be used on the following possible combinations:
    r\equiv3 mod 5, s\equiv 4 mod 5
    r\equiv4 mod 5, s\equiv 2 mod 5
    r\equiv2 mod 5, s\equiv 4 mod 5
    r\equiv3 mod 5, s\equiv 1 mod 5
    r\equiv1 mod 5, s\equiv 3 mod 5

    If r\equiv4 mod 5 s\equiv 1 mod 5 then r+s\equiv4+1\equiv0 mod 5, so 5|a
    The previous approach can also be used on the following
    r\equiv1 mod 5, s\equiv 4 mod 5
    r\equiv3 mod 5, s\equiv 2 mod 5
    r\equiv2 mod 5, s\equiv 3 mod 5

    These cover all possible combinations of equivalences modulo 5

    thus, the product abc will have factors of 4, 5 and 3 and so 60|abc

    Edit: how could I possibly have missed seeing PaulRS's post? :s
    Last edited by badgerigar; December 7th 2007 at 10:23 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pythagorean triples
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: June 20th 2011, 04:58 AM
  2. pythagorean triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: November 23rd 2009, 09:15 PM
  3. Pythagorean triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 17th 2009, 06:06 PM
  4. Pythagorean Triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 8th 2008, 11:18 PM
  5. pythagorean triples
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: November 2nd 2008, 05:28 AM

Search Tags


/mathhelpforum @mathhelpforum