Exercise 3.3.6 ($x^2 + (p+1)/4 \equiv 0 \pmod p$ is not solvable, with $p = 4k+3$)

Question

For any prime $p$ of the form $4k+3$, prove that $x^2 + (p+1)/4 \equiv 0 \pmod p$ is not solvable.

Richard Ganaye · Accepted Answer

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ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} Let $p = 4k+3$ be a prime number. Then $4k+3 \equiv 0 \pmod p$, thus $4(k+1)\equiv 1 \pmod p$. Let $a \in \Z$ be an inverse of $2$ modulo $p$, so that $2  a \equiv 1 \mod p$. Then
$$ k + 1 \equiv a^2 \pmod p.$$
Therefore, using Theorem 3.1(1),
\begin{align*}
\legendre{- \, \frac{p+1}{4}}{p} &\equiv \left(-\frac{p+1}{4} ight)^{\frac{p-1}{2}} \pmod p,
\end{align*}
where 
\begin{align*}
\left(-\frac{p+1}{4} ight)^{\frac{p-1}{2}} &= [-(k+1)]^{2k+1}= - (k+1)^{2k+1}\
&\equiv -\left(a^2ight)^{2k+1} = - a^{4k+2}\
&\equiv -a^{p-1}\
&\equiv -1 \pmod p,
\end{align*}
by Fermat's Theorem.

This shows that the congruence $x^2 + (p+1)/4 \equiv 0 \pmod p$ is not solvable.
\end{proof}

Exercise 3.3.6 ($x^2 + (p+1)/4 \equiv 0 \pmod p$ is not solvable, with $p = 4k+3$)

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Exercise 3.3.6 ($x^2 + (p+1)/4 \equiv 0 \pmod p$ is not solvable, with $p = 4k+3$)

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