What is the proof of the theorem which says: there is a root of the equation $x^{2}+x=1$ in $\mathbb{Z}_{p}$ if and only if $p=5$ or $p\equiv -1\bmod 5$ or $p\equiv 1\bmod 5$.
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There are no solutions in $\mathbb{Z}_2$ by inspection. For any prime $p$ other than $2$, we have that $2$ is a unit in $\mathbb{Z}_p$, so that $$x^2+x\equiv 1\bmod p\iff 4x^2+4x\equiv 4\bmod p\iff (2x+1)^2=4x^2+4x+1\equiv 5\bmod p.$$ Thus, we see that there the equation has solutions in $\mathbb{Z}_p$ if and only if $5$ is a quadratic residue (or $0$) modulo $p$. By quadratic reciprocity $5$ is a quadratic residue modulo $p$ if and only if $p\equiv1\bmod 5$ or $p\equiv 4\equiv-1\bmod 5$, and $5$ is $0$ modulo $p$ if and only if $p=5$.
Zev Chonoles
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Exactly like in $\mathbb R$, quadratic equations in $\mathbb Z_p$ have a solution if and only if the discriminant is a square un $\mathbb Z_p$.
Here we have $\Delta = 5$ so the question now is : does $5$ have a square root in $\mathbb Z_p$ ?
This is answered by quadratic reciprocity.
Damien L
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