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Let $a,n\ \in \mathbb Z$ and suppose that $n>1$ is odd, $n\equiv3\pmod{4}$, and that $\gcd(a,n)=1$.

Prove that if $a^{(n-1)/2}\equiv\pm1\pmod{n}$, then $$\left(\frac{a}{n}\right)\equiv a^{(n-1)/2}\pmod{n}$$

I have no idea how to prove the desired result. I started by noting that $a$ is therefore not a Miller-Rabin Witness by our assumption. Further we know that for $n=2^kq+1$, $k=1$ since $n\equiv3\pmod{4}$. However, putting this all together I just got back to the initial assumption that $a^{(n-1)/2}\equiv\pm1\pmod{n}$.

Should I focus more on the other side of the equation (i.e. the Jacobi Symbol)? I was thinking this but I couldn't figure out anything to do with it other than to break it up into: $$ \left(\frac{a}{n}\right)=\left(\frac{1}{n}\right)\left(\frac{a}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{-1}{n}\right)\left(\frac{a}{n}\right) $$ This didn't seem to take me very far either, however, since $ \left(\frac{1}{n}\right)=1$ and $\left(\frac{-1}{n}\right)=-1$. Any help would be appreciated.

rtybase
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Mjoseph
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1 Answers1

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From $n \equiv 3 \pmod{4}$, we have $\exists q\in\mathbb{N}$ s.t. $n=4q+3$.

Now, let's assume $a^{\frac{n-1}{2}} \equiv \color{blue}{1} \pmod{n}$ then $$a^{2q+1}\equiv 1\pmod{n} \Rightarrow \left(\color{red}{a^{q+1}}\right)^2\equiv a\pmod{n}$$ which means $\left(\frac{a}{n}\right)=\color{blue}{1}$.

Similarly, let's assume $a^{\frac{n-1}{2}} \equiv \color{blue}{-1} \pmod{n}$ then $$a^{2q+1}\equiv -1\pmod{n} \Rightarrow \left(\color{red}{a^{q+1}}\right)^2\equiv -a\pmod{n}$$ which means $\left(\frac{-a}{n}\right)=1$. But $1=\left(\frac{-a}{n}\right)=\left(\frac{-1}{n}\right)\cdot\left(\frac{a}{n}\right)$ and $$\left(\frac{-1}{n}\right)=(-1)^{\frac{n-1}{2}}=(-1)^{2q+1}=-1$$ as a result $\left(\frac{a}{n}\right)=\color{blue}{-1}$.

rtybase
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