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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

Best Answer 

 #1
avatar+20025 
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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

 

laugh

heureka  Aug 11, 2017
 #1
avatar+20025 
+1
Best Answer

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

 

laugh

heureka  Aug 11, 2017

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