Let $n$ be a positive integer and $a$ be an integer such that $a$ is its own inverse modulo $n$. What is the remainder when $a^2$ is divided by $n$?
Let n be a positive integer and a be an integer such that a is its own inverse modulo n.
What is the remainder when a^2 is divided by n?
A modular inverse of an integer a (modulo n) is the integer \(a^{(-1)}\) such that
\(aa^{(-1)}=1 \pmod n\)
So let \(a^{(-1)} = a\), we have:
\(\begin{array}{|rcll|} \hline a\cdot a &=& 1 \pmod n \\ a^2 &=& 1 \pmod n \\ \hline \end{array}\)
The remainder when a2 is divided by n ist 1
Let n be a positive integer and a be an integer such that a is its own inverse modulo n.
What is the remainder when a^2 is divided by n?
A modular inverse of an integer a (modulo n) is the integer \(a^{(-1)}\) such that
\(aa^{(-1)}=1 \pmod n\)
So let \(a^{(-1)} = a\), we have:
\(\begin{array}{|rcll|} \hline a\cdot a &=& 1 \pmod n \\ a^2 &=& 1 \pmod n \\ \hline \end{array}\)
The remainder when a2 is divided by n ist 1