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I need to find all natural solutions to $x^2 + 2y^2 = z^2$

What I tried: I did $\pmod 2$ to the equation receiving $z^2 - x^2 \equiv 0 \pmod 2$.

Then there are two possibilities:

  1. $x^2 \equiv 0 \pmod 2$, $z^2 \equiv 0 \pmod 2$ then $x \equiv 0 \pmod 2$ , $z \equiv 0 \pmod 2$

  2. $x^2 \equiv 1 \pmod 2$ , $z^2 \equiv 1 \pmod 2$ then $x \equiv 1 \pmod 2$ , $z \equiv 1 \pmod 2$

In the first possibility I divide the equation by $2$ infinitely many times and get a contradiction. So I get only the trivial $(0,0,0)$ solution.

In the second one I'm stuck... Any help will be appreciated.

Bart Michels
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user2993422
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1 Answers1

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Every number that is not of the form $4k+2$ can be written as a difference of two squares, hence a solution exists for every even $y$. Just an example, $y=6$. Then $2y^2 = 2^3\cdot 3^2$, so by setting $(z-x)=2\cdot 3$ and $(z+x)=2^2\cdot 3$ we get the solution $(x,y,z)=(3,6,9)$.

Jack D'Aurizio
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