Results 1 to 2 of 2

Math Help - diophantine equation

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    98

    diophantine equation

    Find all solutions in positive integers of each Diophantine equation below.

    a)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by bigb View Post
    Find all solutions in positive integers of each Diophantine equation below.

    a)
    Let x^2 + 2y^2 = z^2 be positive integers.
    We can consider the case when \gcd(x,y,z)=1.
    In such a case we require that x,z \text{ odd } and y \text{ even}.
    Another observation is that \gcd(x,z)=1.

    Given two positive odd integers x<z it turns out that either 4|(z+x) or 4|(z-x).
    These will be our two cases to consider.

    Case 1 z+x \equiv 0 (\bmod 4): We can write 2y^2 = (z+x)(z-x) and so 2(\tfrac{y}{2})^2 = \left( \tfrac{z+x}{2} \right)\left( \tfrac{z-x}{2} \right)
    Notice that, \gcd\left( \tfrac{z+x}{2},\tfrac{z-x}{2} \right) = 1
    Now write, (\tfrac{y}{2})^2 = \left( \tfrac{z+x}{4} \right) \left( \tfrac{z-x}{2} \right) with \gcd \left( \tfrac{z+x}{4}, \tfrac{z-x}{2} \right) = 1
    We have a situation when product of two relatively prime integers is a square, so each integer is a square.
    This gives, \tfrac{z+x}{4} = a^2 \text{ and }\tfrac{z-x}{2} = b^2 \implies z+x = 4a^2 \text{ and }z-x = 2b^2 \text{ with }\gcd(a,b)=1
    Solving this we get, x = 2a^2 - b^2, ~ y=2ab, ~ z = 2a^2 + b^2
    Also, (2a^2 - b^2)^2 + 2(2ab)^2 = (2a^2 + b^2)^2
    This shows that the parametric description of the equations above give us a complete list for solutions.

    Case 2: z-x\equiv 0(\bmod 4): We can write 2y^2 = (z+x)(z-x) and so 2(\tfrac{y}{2})^2 = \left( \tfrac{z+x}{2} \right)\left( \tfrac{z-x}{2} \right)
    Notice that, \gcd\left( \tfrac{z+x}{2},\tfrac{z-x}{2} \right) = 1
    Now write, (\tfrac{y}{2})^2 = \left( \tfrac{z+x}{2} \right) \left( \tfrac{z-x}{4} \right) with \gcd \left( \tfrac{z+x}{2}, \tfrac{z-x}{4} \right) = 1
    This gives, \tfrac{z+x}{2} = a^2 \text{ and }\tfrac{z-x}{4} = b^2 \implies z+x =2a^2 \text{ and }z-x=4b^2 \text{ with }\gcd(a,b)=1
    Solving this we get, x= a^2 - 2b^2, ~ y=2ab, ~ z = a^2 + 2b^2
    Also, (a^2 - 2b^2)^2 + 2(2ab)^2 = (a^2 + 2b^2)^2
    This shows that the parametric description of the equations above give us a complete list for solutions.

    Case 1 and Case 2 can be combined by saying: x=|a^2 - 2b^2|, ~ y=2ab, ~ z = a^2 + 2b^2 as a complete list
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. yet another Diophantine equation
    Posted in the Number Theory Forum
    Replies: 12
    Last Post: September 7th 2011, 07:48 AM
  2. diophantine equation
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 1st 2011, 08:19 AM
  3. Diophantine equation
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: January 16th 2011, 02:06 PM
  4. Diophantine Equation
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 8th 2010, 08:01 PM
  5. diophantine equation
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: June 3rd 2009, 06:08 AM

Search Tags


/mathhelpforum @mathhelpforum