Let $a$ be a quadratic residue $\mod{43}$. Show that $a^{11}$ is a ‘square root’ of $a$. There is another square root of $a$, what is it?
A thorough explanation for begginers is appreciated.
Let $a$ be a quadratic residue $\mod{43}$. Show that $a^{11}$ is a ‘square root’ of $a$. There is another square root of $a$, what is it?
A thorough explanation for begginers is appreciated.
As $a$ is a quadratic residue modulo $43$, then $\exists x$, s.t. $x^2 \equiv a \pmod {43}$
If $a^{11}$ is the square root of $a$ in $\mathbb{Z}^{*}_{43}$ then we have that $a \equiv a^{22} \pmod{43} \iff x^2 \equiv x^{44} \pmod{43}$. But the follwoing is true because of the Fermat's Little Theorem, as $x^{43} \equiv x \pmod{43}$