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Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.

Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?

Thanks for any help.

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    While it is true that any solution can be expressed with $x=r^2-s^2$, $y=2rs$, and $z^2=r^2+s^2$ (note the square in $z$) because $x$, $y$, and $z^2$ are a primitive pythagorean triple, not every pair of values of $r$ and $s$ (with $r\gt s$, $\gcd(r,s)=0$, and $r$ and $s$ of opposite parity) will yield a solution, since you need $r^2+s^2$ to be a square. But you can then implement the formula *again* to find solutions to $z^2=r^2+s^2$, and then...2012-07-18
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    Let $x+yi = (a+bi)^4$2012-07-18
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    What do you mean with "$(x,y,z)=1$"?2012-07-18
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    @celtschk: It means $x,y,$ and $z$ are relatively prime.2012-07-18

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