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Let $a,a_1,a_2$ be integers and $b,b_1,b_2$ be odd positive integers.

Prove $(a_1a_2 / b) = (a_1/b)(a_2/b)$

Prove $(a / b_1b_2) = (a/b_1)(a/b_2)$ I need to prove these 2, i know both deal with the jacobi symbol but stuck.

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  • $\begingroup$ Oh, dear! Please go to the FAQ section and check there how to type mathematics with LaTeX in this site, otherwise if is hard and very uncomfortable to read mathematics in ASCII. $\endgroup$ – DonAntonio Nov 28 '12 at 4:44
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We change names slightly, for no very good reason. And again for no good reason, we do the second question first.

Recall that if $m$ is odd, and has prime power factorization $p_1^{e_1}\cdots p_k^{e_k}$, then by the definition of the Jacobi symbol, $$(x/m)=\prod_{i=1}^k (x/p_i)^{e_i}.\tag{$1$}$$ (The symbols on the right are Legendre symbols. Unfortunately, the same notation is used for each.)

The formula remains correct if we include in the above product some primes $p$ that actually don't divide $m$. For $(x/p)^0=1$, so does not affect the above product.

Now consider $(x/bc)$. Let $b=\prod_{i=1}^k p_i^{e_i}$ and $c=\prod_{i=1}^k p_i^{f_i}$. (Remember that some $e_i$ or $f_i$ can be $0$.) Then $$bc=\prod_{i=1}^k p_i^{e_i+f_i},$$ and now the result follows immediately from $(1)$.

For the first question, again use the basic defining Equation $(1)$. We want to show that $(x/m)(y/m)=(xy,m)$. To compute $(x/m)$, we find a product of terms $(x/p_i)^{e_i}$, where the $(x/p_i)$ are Legendre symbols. We compute $(y/m)$ and $(xy/m)$ is a similar way.

However, for Legendre symbols, the relationship $(xy/p)=(x/p)(y/p)$ holds. I am reasonably sure that this has already been done in your course. The result now follows immediately from $(1)$.

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