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# 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

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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 $n$?

RektTheNoob  Aug 10, 2017

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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

heureka  Aug 11, 2017
#1
+19496
+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

heureka  Aug 11, 2017