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How do I use Quadratic Reciprocity to compute $\left( \frac{11}{1729} \right)$?

Attempt: $\left( \frac{11}{1729} \right) = \left( \frac{p}{q} \right)$ where $p,q$ are primes with $p = 3 \pmod 4$, $q = 1 \pmod 4$. But the quadratic reciprocity theorem from my textbook only gives the answer for cases $p,q$ both $1 \pmod 4$ or both $3 \pmod 4$. So I'm stuck.

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    $\begingroup$ I meant to point out that the composite nature of $1729$ means we are actually dealing here with quadratic reciprocity of the Jacobi symbol, but this is substantially dealt with in the Comments on @Timbuc's Answer. $\endgroup$
    – hardmath
    Mar 6, 2015 at 17:34

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since $\;1729=2\pmod{11}\;$ :

$$\binom{11}{1729}=\binom{11}2=-1$$

as $\;11\neq\pm1\pmod 8\;$ .

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  • $\begingroup$ I don't see how $\dbinom{11}{1729}$ is defined since $1729$ is not prime. Even if you think of it as the Jacobi symbol, I don't see why it should be multiplicative for the lower integer. $\endgroup$
    – Bernard
    Mar 6, 2015 at 17:06
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    $\begingroup$ @Bernard, by definition of the Jacobi's Symbol:$$1729=7\cdot13\cdot19\implies\binom{11}{1729}=\binom{11}7 \binom{11}{13}\binom{11}{19}=\left(-\binom7{11}\right)\binom2{11}\left(-\binom8{11}\right)=-1$$ $\endgroup$
    – Timbuc
    Mar 6, 2015 at 17:11
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    $\begingroup$ I think you should have given some details for the O.P. $\endgroup$
    – Bernard
    Mar 6, 2015 at 17:13
  • $\begingroup$ @Bernard I think that if the OP understands his own question this must suffice. Anyway, with theee comments things should come around easier, and if he has any doubts he can ask. Thanks. $\endgroup$
    – Timbuc
    Mar 6, 2015 at 17:14

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