Legendre calculation passing through Jaccobi We know that Jacobi's symbol does not guaranty that if $(\frac{a}{b}) = 1$ then $a$ is a quadric remander mod $b$.
We also know that when calculating Legendre symbol you can use quadratic reciprocity like this $(\frac{a}{p}) = \pm (\frac{p}{a})$ even if you get a non-prime denominator.
We were told that the result does not lose it's validity if you start the calculation with a prime denominator. we can understand why?
And if that is true, why does a calculation starting with a composite denominator does not gain validity if we "pass through" a prime denominator.
Thanks!
 A: If you start with a prime in the denominator, for instance $\newcommand{\kron}[2]{\left( \frac{#1}{#2} \right)}$ $\kron{4}{7}$, or more generally $\kron{a}{p}$, then this is exactly the Legendre symbol, defined to be $1$ if the numerator is a square mod the denominator and $-1$ if the numerator is not a square mod the denominator (and $0$ if the denominator divides the numerator). Many further details about the Legendre symbol can be picked up in any book on elementary number theory (such as Silverman's Friendly Intro to Number Theory, which I have used as a resource in courses I've taught, or K. Rosen's Elementary Number Theory and its Applications, which I first learned elementary number theory from).
However, simply passing through a prime denominator is insufficient. It suffices to produce a counterexample. For instance, $\kron{3}{35} = 1$, but $3$ is not a square mod $35$. Yet you can compute this through $\kron{3}{35} = \kron{3}{5} \kron{3}{7}$, which have primes in the denominators --- but both symbols on the right are $-1$.
