# Square Root Modulo Composite

• May 20th 2010, 03:21 AM
EinStone
Square 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 AM
Chris L T521
Quote:

Originally Posted by EinStone
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.

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 AM
EinStone
Quote:

Originally Posted by Chris L T521
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.

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

Quote:

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?
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 $\displaystyle n-1\equiv n-a^{\varphi(n)/2}\pmod{n}$

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.

Quote:

What do you mean increase n, I mean n is fixed??
I mean as we pick larger values of $\displaystyle n$, 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 $\displaystyle x^2\equiv 1\pmod{p_i}$ for each prime (power) divisor $\displaystyle p_i$. Does this yield at least four solutions modulo $\displaystyle n$?
• Jun 2nd 2010, 10:17 AM
chiph588@
Quote:

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

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

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

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 PM
chiph588@
Quote:

Originally Posted by EinStone
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.

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.

Quote:

Originally Posted by roninpro
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 $\displaystyle n$?

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.