Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$[\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).

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.