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We have a Diophanite equation $$x^2+2y^2=p.$$ Suppose $p$ is prime, and there exists a solution $(a,b)\in\mathbb{Z}\times\mathbb{Z}$ to the Diophanite equation. Prove that there exists $z\in\mathbb{Z}$ for which $z^2\equiv-2\bmod{p}$.

Hint: Show that $p\nmid{b}$, and hence $\overline{b}\in(\mathbb{Z}/p\mathbb{Z})^\times$; then use the multiplicative inverse of $\overline{b}$ modulo $p$ to construct $z$.

I already proved $\overline{b}\in(\mathbb{Z}/p\mathbb{Z})^\times$, but I'm struggling to use the multiplicative inverse of $\overline{b}$ to complete this proof. I would really appreciate any hints/suggestions.

Bill Dubuque
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Hana
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  • What do you mean by "finding" the multiplicative inverse? $\overline b$ is already the multiplicative inverse of $b$ modulo $p$, and you can use that symbol in further arguments. – Greg Martin Mar 30 '21 at 06:10
  • I apologize for the confusion:(. I changed the word from "find" to "use." I just did not know how to use $\overline{b}$ to complete the proof (i.e. show the existence of $z\in\mathbb{Z}$ for which $z^2\equiv-2\bmod{p}$). – Hana Mar 30 '21 at 06:17
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    Would the manipulation $a^2 \equiv -2b^2 \Leftrightarrow (a \bar{b})^2 \equiv -2$ suffice? – Bimo Adityarahman Mar 30 '21 at 06:37
  • That really helps! Thank you so much. – Hana Mar 30 '21 at 06:51

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