Exercise 3.3.4 (Is $x^4 \equiv 25 \pmod {1013}$ solvable?)

Question

Determine whether $x^4 \equiv 25 \pmod {1013}$ is solvable, given that $1013$ is a prime.

Richard Ganaye · Accepted Answer

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

\begin{proof} 
Since $p = 1013$ is a prime number,
$$x^4 \equiv 25 \pmod {1013} \iff x^2 \equiv 5 \pmod{1013} 	ext{ or } x^2 \equiv -5 \pmod{1013}.$$
Moreover, $\legendre{-5}{1013} = \legendre{5}{1013} = -1$, therefore the congruence $x^4 \equiv 25 \pmod {1013}$ is not solvable.

(Alternatively, since $4 \mid p-1$, and $25^{(p-1)/4}  = 5^{(p-1)/2}  \equiv \legendre{5}{1013} = -1 \pmod p$, by Theorem 2.37, $25$ is not a fourth power modulo $1013$.)
\end{proof}

Exercise 3.3.4 (Is $x^4 \equiv 25 \pmod {1013}$ solvable?)

Answers

Comments

Exercise 3.3.4 (Is $x^4 \equiv 25 \pmod {1013}$ solvable?)

Answers

Comments

Add answer