Number of elements in a multiplicative group

Question

Given $n$ the square of a prime odd number. How many elements $\bar x ∈$ $Z^*_n$ with $\bar x ^2$ = $\bar 1$?

Given $n$ the product of two distinct prime numbers. How many elements $\bar x ∈$ $Z^*_n$ with $\bar x ^2$ = $\bar 1$?

Not knowing the value of $n$, how can I find out that cardinality? Is there any theorem that I should follow to induce $n$?

I think that if you work it out by hand for a few specific $n$, you'll see a pattern and realize the answer to your question. — Mike Pierce, Jul 17 '18 at 14:50
Euler's generalization of Fermat's Little Theorem is certainly relevant here. — hardmath, Jul 17 '18 at 14:55
@hardmath I know there is a relation between them but actually don't know how to connect the points. — CptPackage, Jul 17 '18 at 15:59
I'm going to suggest you check the definition of "multiplicative group" $\mathbb Z_n^*$ and the well-known characterization of which ones are cyclic. You can also approach this problem using logic of the Chinese Remainder Thm., noting that $x^2 \equiv 1 \bmod p$ is a polynomial equation over a field when $p$ is prime. — hardmath, Jul 17 '18 at 16:54

Sean Haight · Answer 1 · 2019-06-13T04:41:38.343

For the first question: Let $n = p^2$. We're interested in $x$ such that $x^1 \equiv 1 \pmod {p^2}$. That is $p^2 | x^2 - 1$. We see that this is then equivalent to $p^2 | (x - 1)(x + 1)$. We have three cases:

$p^2|(x - 1)$: In this case $x \equiv 1 \pmod p^2$.

$p^2|(x + 1)$: In this case $x \equiv p^2-1 \pmod p^2$.

Thus if $n = p^2$ for an odd prime there are exactly two solutions to $x^2 \equiv 1 \pmod n$, namely $1$ and $p^2 - 1$.

Next suppose $n = pq$ for distinct primes $p$ and $q$. Again we can arrive at the equation $pq|(x + 1)(x - 1)$. We then have four cases:

$p|(x + 1)$, $q|(x - 1)$: In this case we have $x + 1 = kp$ and $x - 1 = mq$. That is $x \equiv 1 \pmod q$ and $x \equiv -1 \pmod p$. Note that by the Chinese Remainder Theorem, this has a unique solution modulo $pq$.

$p|(x - 1)$, $q|(x + 1)$: As above this congruence has a unique solution modulo $pq$ by the Chinese Remainder Theorem.

$pq|(x + 1)$: In this case $x \equiv pq - 1 \pmod{pq}$

$pq|(x - 1)$: In this case $x \equiv 1 \pmod{pq}$

Next note that one $x$ cannot satisfy any two of the cases above. This is because if such an $x$ did exist then $p|(x + 1)$, $p|(x - 1)$ and $q|(x + 1)$ and $q|(x - 1)$. As in part one, this would for $p = q = 2$, which is a contradiction. Thus there are exactly 4 solutions to $x^2 \equiv 1 \pmod n$ if $n$ is the product of two distinct primes.

score 0 · Answer 2 · answered Jul 17 '18 at 17:42

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$U(p^2)$ is cyclic and so $x^2=1$ has exactly two solutions.

$U(pq) \cong U(p) \times U(q)$ is a product of two cyclic groups and so $x^2=1$ has exactly four solutions.

answered Jul 17 '18 at 17:42

lhf

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Number of elements in a multiplicative group

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