Solve $x^2 + x + 47 ≡ 0 \pmod {7^3}$.
I don't know how to go about doing this question. I've tried completing the square and other routes but I always seem to end up with horrible answers for $x$ involving surds.
Any help or hints are appreciated.
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Let $y=2x+1$, then $x^2+x+47\equiv 0$ is equivalent to $y^2+187\equiv 0$. For $p=7$ this is answered by quadratic reciprocity; for powers we could apply Hensel's Lemma. Actually Qiaochu Yuan has answered the question in general here on MSE.
Dietrich Burde
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That discriminant of binary forms has $x^2 + xy + 47 y^2$ and $7 x^2 + 3 xy + 7 y^2,$ only; one per genus. Which means that $x^2 + xy + 47 y^2$ primitively represents $49$ as $1 + 1 + 47,$ but also means it primitively represents $2401,$ with $ x = 43, y = 3.$ Then $3$ is invertible $\pmod {7^3}$ and $\pmod {7^4}$ to get back to $x^2 + x + 47$ – Will Jagy Oct 21 '16 at 17:33
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The definition of Gauss-Dirichlet composition on page 49 of Cox, here composing the principal form with itself, gives $x = -46, y = 3$ for $x^2 + xy + 47 y^2 = 49^2.$ However, we are free to replace $(x,y)$ by $(x+y, -y)$ or $(-x-y,y)$ giving $(43,3).$ Well. – Will Jagy Oct 21 '16 at 17:59