I am reading the proof the for odd prime $p$, $$ \left ( \frac{-1}{p} \right)_2 = (-1)^{\frac{p-1}{2}} = \begin{cases} 1 \hspace{2mm} \text{for} \hspace{2mm} p \equiv 1 \operatorname{mod} 4 \\ -1 \hspace{2mm} \text{for} \hspace{2mm} p \equiv 3 \operatorname{mod} 4 \end{cases}$$

The proof states that if $-1$ is a square mod $p$, then a square root of it has order $4$ in $(\mathbb{Z}/p)^{\times}$. That is, if $-1 = b^2 \operatorname{mod} p$, then one of the possible square roots of $b$ generates a subgroup of order $4$ in $(\mathbb{Z}/p)^{\times}$.

Why is this true? (Kinda makes sense if we square both sides of the equation but I am not sure this is allowed)


No, if $-1\equiv b^2\bmod p$, then $b$ itself generates a subgroup of order $4$ in $(\mathbb{Z}/p\mathbb{Z})^\times$, namely $$\langle b\rangle\;=\;\{b^0=1,\;\;b^1=b,\;\;b^2=-1,\;\;b^3=-b\}$$ After that, we get $b^4=(-b)\cdot b=(-1)\cdot (-1)=1$. Therefore $\langle b\rangle$ has $4$ elements.

If $b$ has a square root (which need not be the case) then it would generate a subgroup of order $8$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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