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It is well known that any positive odd integer can be written as $x^2+2y^2+4z^2$ with $x,y,z\in\mathbb Z$.

Question 1. Whether for any odd integer $n>93$ there are $x,y,z\in\mathbb Z$ such that $7n=x^2+2y^2+4z^2$ but $x^2\equiv y^2\equiv z^2\pmod 7$ fails?

Question 2. Whether for any odd integer $n>213$ there are $x,y,z\in\mathbb Z$ such that $7n=x^2+2y^2+4z^2$ and $x^2\equiv y^2\equiv z^2\pmod 7$?

My computation suggests that both questions might have positive answers, but I don't know how to prove this.

Your helpful comments are welcome!

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  • $\begingroup$ Note that $1^2+2\times3^2+4\times2^2=35\equiv0\pmod 7$. $\endgroup$
    – Zhi-Wei Sun
    Commented Oct 21, 2020 at 13:30

1 Answer 1

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Let $f(x,y,z):=x^2+2y^2+4z^2$. Then the following automorph of $f$ may help you to solve the problem.

\begin{equation} f(x,y,z)=f(\frac{x+8y+8z}{7},\frac{-4x+3y-4z}{7},\frac{-2x-2y+5z}{7}). \end{equation}

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    $\begingroup$ If $y\equiv-3x\pmod7$ and $z\equiv2x\pmod 7$, then $x+y+z\equiv 0\pmod 7$ and so your identity is helpful to Question 2. If $x^2\equiv y^2\equiv z^2\not\equiv0\pmod 7$, then $x+y+z\not\equiv0\pmod 7$ and hence the identity offers no help to Question 1. $\endgroup$
    – Zhi-Wei Sun
    Commented Oct 22, 2020 at 0:20

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