Show that for any integer $n=1,..,32$ such that $(n,33) = 1$, $\left(\frac{n}{33}\right) = \left(\frac{33-n}{33}\right)$

Question

Is there a way to verify this without going through all values of n?

Answer 1

Yes, there is a way. The key fact is that $3$ and $11$ are both of the form $4k+3$.

The Jacobi symbol $(a/33)$ is defined as $(a/3)(a/11)$, where $(a/3)$ and $(a/11)$ are Legendre symbols.

Note that $((33-a)/3)=(-a/3)$ and $((33-a)/11)=(-a/11)$ by properties of the Legendre symbol. So we need to show that $(a/3)(a/11)=(-a/3)(-a/11)$.

But $(-a/3)=(-1/3)(a/3)=-(a/3)$. For $(-1/3)=-1$, since $3$ is of the shape $4k+3$. Similarly, $(-a/11)=-(a/11)$, since $11$ is of the shape $4k+3$. The result follows.

Remark: We travelled through properties of the Legendre symbol. But we could have worked purely with standard properties of the Jacobi symbol, if those have already been derived. For $((33-a)/33)=(-a/33)=(-1/33)(a/33)$. But $(-1/33)=(-1/3)(-1/11)=(-1)(-1)=1$.