Square Root Modulo Composite

• May 20th 2010, 03:21 AM
EinStone
Square Root Modulo Composite
Show that if $n$ is composite, $n \neq 2p$ (for prime p), then there at least 4 different residues modulo n, whose squares equal 1.
• May 20th 2010, 05:51 AM
Chris L T521
Quote:

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

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

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

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

Does this make sense?
• May 20th 2010, 08:35 AM
EinStone
Quote:

Originally Posted by Chris L T521
This one generates a residue different from $1$ since we required $n\neq 2p$ and it can't generate $n-1$; if it did, then $n-1\equiv n-a^{\varphi(n)/2}\pmod{n}\implies a^{\varphi(n)/2}\equiv 1\pmod{n}$, which isn't possible given the restrictions we made on the problem.

Can you explain this again, esepcially how you get $n-1\equiv n-a^{\varphi(n)/2}\pmod{n}$

Quote:

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

Does this make sense?
What do you mean increase n, I mean n is fixed??
• May 20th 2010, 08:42 AM
Chris L T521
Quote:

Originally Posted by EinStone
Can you explain this again, esepcially how you get $n-1\equiv n-a^{\varphi(n)/2}\pmod{n}$

I was claiming that $n-1$ couldn't be generated from $n-a^{\varphi(n)/2}$. To show that, suppose that they were equivalent. Then $n-1\equiv n-a^{\varphi(n)/2}\pmod{n}\implies 1\equiv a^{\varphi(n)/2}\pmod{n}$, which can not happen.

Quote:

What do you mean increase n, I mean n is fixed??
I mean as we pick larger values of $n$, the number of residues that satisfy the property of interest will be more than 4.
• May 20th 2010, 09:42 AM
roninpro
I was actually thinking that you could use the Chinese Remainder Theorem. Just look at $x^2\equiv 1\pmod{p_i}$ for each prime (power) divisor $p_i$. Does this yield at least four solutions modulo $n$?
• Jun 2nd 2010, 10:17 AM
chiph588@
Quote:

Originally Posted by Chris L T521
$\implies 1\equiv a^{\varphi(n)/2}\pmod{n}$, which can not happen.

$(2k+1)^{\phi(8)/2}=(2k+1)^2=4k(k+1)+1\equiv1\bmod{8}$
• Jun 2nd 2010, 01:43 PM
chiph588@
Quote:

Originally Posted by chiph588@
$(2k+1)^{\phi(8)/2}=(2k+1)^2=4k(k+1)+1\equiv1\bmod{8}$

In fact with a little work one can show $a^{\phi(m)/2}\equiv1\bmod{m}$ when $m$ is not of the form $1,2,4,p^a,2p^a$, where $p$ is a prime.

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

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I think the best way to approach this problem would be to show there exists $a$ such that $|a|\not\equiv1\bmod{n}$ and $a\equiv a^{-1}\bmod{n}$.
• Jun 2nd 2010, 03:59 PM
chiph588@
Quote:

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

I believe this statement is false. Consider $18=2\cdot3^2\neq2p$ for a prime $p$

$\begin{tabular}{|c c | c c|}\hline
a & a squared & a & a squared \\\hline
1 & 1 & 10 & 10\\
2 & 4 & 11 & 13\\
3 & 9 & 12 & 0\\
4 & 16 & 13 & 7\\
5 & 7 & 14 & 16\\
6 & 0 & 15 & 9\\
7 & 13 & 16 & 4\\
8 & 10 & 17 & 1\\
9 & 9 & & \\\hline
\end{tabular}$

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

Quote:

Originally Posted by roninpro
I was actually thinking that you could use the Chinese Remainder Theorem. Just look at $x^2\equiv 1\pmod{p_i}$ for each prime (power) divisor $p_i$. Does this yield at least four solutions modulo $n$?

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