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$[\frac{109}{1925} ]=[\frac{109}{5} ]^2[\frac{109}{7} ][\frac{109}{11} ] = [\frac{4}{5} ]^2[\frac{4}{7} ][\frac{-1}{11} ] = (?)^2[\frac{2^2}{7} ][\frac{-1}{11} ] = (?)^2\cdot 1 \cdot (-1)^{\frac{11-1}{2}}=1 \cdot 1 \cdot (-1) = -1 $ So how you show that $[\frac{a}{n} ]^2=1$, where $a \in \mathbb{Z}$ and $n$ is odd integer?( $[\frac{a}{n} ]$ is Jacobi symbol and when $p$ is prime, then $[\frac{a}{p} ]$ is Legendre symbol).

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I'm not sure what you are mising. $-1$ is not a square mod $11$ and $4$ is a square mod whatever. –  Hagen von Eitzen Jun 4 '13 at 17:19
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up vote 1 down vote accepted

The Jacobi symbol is defined as a product of Legendre symbols. Legendre symbols are $\pm 1$, so their square is always $1$. The same is therefore true of Jacobi symbols.

In particular, in the calculation, there was no reason to transform $(109/5)^2$ to $(4/5)^2$. The result is correct, but the fact that $(109/5)^2=1$ requires no calculation.

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