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    diophantine equation

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

    a)
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    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
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