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.

• 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. Mar 6, 2015 at 17:34

since $\;1729=2\pmod{11}\;$ :
$$\binom{11}{1729}=\binom{11}2=-1$$
as $\;11\neq\pm1\pmod 8\;$ .
• 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. Mar 6, 2015 at 17:06
• @Bernard, by definition of the Jacobi's Symbol:$$1729=7\cdot13\cdot19\implies\binom{11}{1729}=\binom{11}7 \binom{11}{13}\binom{11}{19}=\left(-\binom7{11}\right)\binom2{11}\left(-\binom8{11}\right)=-1$$ Mar 6, 2015 at 17:11