Show that if $\displaystyle n$ is composite, $\displaystyle n \neq 2p$ (for prime p), then there at least 4 different residues modulo n, whose squares equal 1.

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- May 20th 2010, 03:21 AMEinStoneSquare Root Modulo Composite
Show that if $\displaystyle n$ is composite, $\displaystyle n \neq 2p$ (for prime p), then there at least 4 different residues modulo n, whose squares equal 1.

- May 20th 2010, 05:51 AMChris L T521
To make this easy to follow, suppose we're dealing with $\displaystyle \{0,1,\ldots,n-1\}$, the standard residue system modulo $\displaystyle n$. Given any $\displaystyle n$ composite, we're always guaranteed two residues with this property since $\displaystyle 1^2\equiv 1\pmod{n}$ and $\displaystyle (n-1)^2\equiv n^2-2n+1\equiv 1\pmod{n}$.

Let $\displaystyle a$ be a residue modulo $\displaystyle n$ (where $\displaystyle n\neq 2p$) with $\displaystyle 1<a<n-1$ and let $\displaystyle (a,n)=1$. Then $\displaystyle a^{\varphi(n)}\equiv 1\pmod{n}$ by Euler's Theorem. Keeping in mind that $\displaystyle 2\mid \varphi(n)$ when $\displaystyle n>2$, we see that $\displaystyle a^{\varphi(n)}\equiv \left(a^{\varphi(n)/2}\right)^2\equiv 1\pmod{n}$. Thus, $\displaystyle a^{\varphi(n)/2}$ is a residue that satisfies this property. Consequently, $\displaystyle -a^{\varphi(n)/2}\equiv n-a^{\varphi(n)/2}\pmod{n}$ is also another residue. This one generates a residue different from $\displaystyle 1$ since we required $\displaystyle n\neq 2p$ and it can't generate $\displaystyle n-1$; if it did, then $\displaystyle n-1\equiv n-a^{\varphi(n)/2}\pmod{n}\implies a^{\varphi(n)/2}\equiv 1\pmod{n}$, which isn't possible given the restrictions we made on the problem.

This shows that we are guaranteed four different residues modulo $\displaystyle n$ with this property. We then allow for the possibility of more residues as we allow $\displaystyle n$ to increase.

Does this make sense? - May 20th 2010, 08:35 AMEinStone
Can you explain this again, esepcially how you get $\displaystyle n-1\equiv n-a^{\varphi(n)/2}\pmod{n}$

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This shows that we are guaranteed four different residues modulo $\displaystyle n$ with this property. We then allow for the possibility of more residues as we allow $\displaystyle n$ to increase.

Does this make sense?

- May 20th 2010, 08:42 AMChris L T521
I was claiming that $\displaystyle n-1$ couldn't be generated from $\displaystyle n-a^{\varphi(n)/2}$. To show that, suppose that they were equivalent. Then $\displaystyle n-1\equiv n-a^{\varphi(n)/2}\pmod{n}\implies 1\equiv a^{\varphi(n)/2}\pmod{n}$, which can not happen.

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What do you mean increase n, I mean n is fixed??

- May 20th 2010, 09:42 AMroninpro
I was actually thinking that you could use the Chinese Remainder Theorem. Just look at $\displaystyle x^2\equiv 1\pmod{p_i}$ for each prime (power) divisor $\displaystyle p_i$. Does this yield at least four solutions modulo $\displaystyle n$?

- Jun 2nd 2010, 10:17 AMchiph588@
- Jun 2nd 2010, 01:43 PMchiph588@
In fact with a little work one can show $\displaystyle a^{\phi(m)/2}\equiv1\bmod{m} $ when $\displaystyle m $ is not of the form $\displaystyle 1,2,4,p^a,2p^a $, where $\displaystyle p $ is a prime.

So unfortunately I believe your proof (for most $\displaystyle n $) only finds $\displaystyle 2 $ such numbers, namely $\displaystyle \{1,n-1\} $, where $\displaystyle a^2\equiv1\bmod{n} $.

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I think the best way to approach this problem would be to show there exists $\displaystyle a $ such that $\displaystyle |a|\not\equiv1\bmod{n} $ and $\displaystyle a\equiv a^{-1}\bmod{n} $. - Jun 2nd 2010, 03:59 PMchiph588@
I believe this statement is false. Consider $\displaystyle 18=2\cdot3^2\neq2p $ for a prime $\displaystyle p $

$\displaystyle \begin{tabular}{|c c | c c|}\hline

a & a squared & a & a squared \\\hline

1 & 1 & 10 & 10\\

2 & 4 & 11 & 13\\

3 & 9 & 12 & 0\\

4 & 16 & 13 & 7\\

5 & 7 & 14 & 16\\

6 & 0 & 15 & 9\\

7 & 13 & 16 & 4\\

8 & 10 & 17 & 1\\

9 & 9 & & \\\hline

\end{tabular} $

Only $\displaystyle (\pm1)^2\equiv1\bmod{18} $, thus a counterexample.

Using this argument in a slightly altered way for my defense, my guess is this theorem is true when $\displaystyle n\neq2p^a $ where $\displaystyle p $ is an odd prime.