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Find number of Integer: $x^2+y^2=12z^2$(1)

Let $x=6u$ and $y=6v$

(1)$\Rightarrow 36u^2+36v^2=12z^2$

$\Leftrightarrow 3(u^2+v^2)=z^2$

So, $z^2 \vdots 3 \Rightarrow z \vdots3 \Rightarrow GCD(x,y,z)=3 $

Equation (1): No solution.

Everyone help me, check and give me some ways solution?

Bill Dubuque
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Hiếu Lê
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    Please edit to include your efforts. Also, note that the question in the header isn't the same as the one in the body. – lulu May 27 '24 at 16:12
  • Welcome to Math SE. FYI, we can show that both $x$ and $y$ must have the same parity. This means they must both be even because if they are both odd then $x^2$ and $y^2$ each leave a remainder of $1$ when divided by $4$, so the LHS has only $1$ factor of $2$, but the RHS has at least $2$ factors of $2$. Thus, $x=2x_1$ and $y=2y_1$, giving $x_1^2+y_1^2=3z^2$, which is answered in the duplicate question. Note I found it using an Approach0 search. – John Omielan May 27 '24 at 16:46
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    If you simply let $r=2z$, the equation is recast as $x^2+y^2=3r^2$, and you get directly to the same kind of equation as in John Omielan's comment. – Keith Backman May 27 '24 at 20:19

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