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Math Help - A Diophantine Problem

  1. #1
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    A Diophantine Problem

    What is the general solution of x^4 + y^2 = z^2 ?

    We know that primitive Pythagorean triples correspond to the triples (2st, s^2 - t^2, s^2 + t^2), where gcd (s,t)=1. So i separate the problem into two cases:

    i) x is even : i found the general solution as
    (sqrt(2st), s^2 - t^2, s^2 + t^2).

    ii) y is even : i found the general solution as
    (sqrt(s^2 - t^2), 2st, s^2 + t^2).

    Is this solution true? Or in other words, do these triples produce every solution of this equation?
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  2. #2
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    Case 1: x is even. Then we know x^2 = 2st , y = s^2 - t^2, z=s^2+t^2 where \gcd(s,t)=1 and the parity is different. It is safe to assume that s is even (because the other way around everything is symettric except for y but that not matter because we are squaring it, and thereby removing the negative sign). But x=2x'. Which means, 4(x')^2 = 2st \implies (x')^2 = (s/2)t (remember s is even). Since \gcd(s/2,t)=1 it means s/2=a^2\implies s = 2a^2 and t=b^2. Thus, x=2ab, y=4a^2 - b^2, z=4a^2+b^2 . For arbitrary relatively prime integers a,b.

    Case 2: x is odd. Then we know x^2 = s^2 - t^2, y=2st, z=s^2+t^2. Thus, x^2 + t^2 = s^2. This forces t to be even. And so, x=a^2 - b^2, t=2ab, s=a^2+b^2. Thus, x=a^2-b^2, y=4ab(a^2+b^2), z=(a^2+b^2) + 4a^2b^2.
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  3. #3
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    Thanks for the help ThePerfectHacker. Can we use the same strategy for the equations x^n + y^2 = z^2, where n>2? Or for which n we can apply the same procedure?

    Note: I have noticed a small typo in the solution:z must be (a^2 + b^2)^2 + 4*((ab)^2).
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  4. #4
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    Quote Originally Posted by enrique View Post
    Thanks for the help ThePerfectHacker. Can we use the same strategy for the equations x^n + y^2 = z^2, where n>2? Or for which n we can apply the same procedure?
    I do not think so. Maybe if n is a power of two.
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