0

I am trying to solve the following problem:

Show that for each $m\in \mathbb{N}$, there exists $n\in \mathbb{N}$ such that $x^2\equiv 1 \pmod n$ has at least $m$ solutions?

I figured that this is equivalent to showing that there is a sequence $n_1,n_2,n_3,\dots$ such that

$$x^2\equiv 1 \pmod {n_1}\qquad x^2\equiv 1 \pmod {n_2}\qquad x^2\equiv 1 \pmod {n_3} \qquad\dots$$

Has an increasing number of solutions. By computation, I figured that this sequence could be

$$n_1=2\qquad n_2 =2\cdot 3\qquad n_3 = 2\cdot 3 \cdot 5\qquad n_4 = 2\cdot 3 \cdot 5 \cdot 7 \qquad \dots $$

Now comes my trouble: I can't figure out why "adding" another prime in each step of the sequence makes more solutions appear. Can you help me?

Red Banana
  • 23,956
  • 20
  • 91
  • 192
  • 1
    Use the Chinese remainder theorem. By the way, you can remove the factor of $2$, it doesn't help at all. – Qiaochu Yuan Dec 09 '20 at 06:24
  • @QiaochuYuan I thought about the CRT but it's still not clear to me. Could you elaborate? – Red Banana Dec 09 '20 at 06:28
  • 2
    If $p$ is an odd prime then the solutions to $x^2 \equiv 1 \bmod p$ are $x \equiv \pm 1 \bmod p$. So if $n$ is a product of distinct odd primes what does CRT tell you the possible solutions are? You can try working this out explicitly for $3 \cdot 5 = 15$ as a warmup. – Qiaochu Yuan Dec 09 '20 at 06:32
  • @QiaochuYuan I tried it but I am a bit confused - I guess am supposed to solve:

    $$x^2\equiv 1 \pmod 3\ x^2\equiv 1 \pmod 5$$

    Using the CRT, right? But the CRT applies only to linear equations, no? I tried to rewrite as:

    $$x\equiv x^{-1} \pmod 3\ x\equiv x^{-1} \pmod 5$$

    But the result doesn't seem to make sense.

     – Red Banana Dec 09 '20 at 07:07
  • 1
    CRT can be used to solve polynomial equations. The conclusion here is that $x \equiv \pm 1 \bmod 3$ and $x \equiv \pm 1 \bmod 5$; this gives four different possibilities for the value of $x \bmod 15$ by combining the two different choices of sign modulo each prime, which explicitly gives $x \equiv 1, -1, 4, -4 \bmod 15$. These are all the solutions to $x^2 \equiv 1 \bmod 15$ and this generalizes. – Qiaochu Yuan Dec 09 '20 at 07:13
  • 1
    @QiaochuYuan I got a bit bugged that CRT can be used to solve polynomial equations, I think I figured out why this is true. We write:

    $$x^2\equiv 1 \pmod 3\x^2\equiv 1 \pmod 5 $$

    From this we get:

    $$(x+1)(x-1)\equiv 0 \pmod 3\(x+1)(x-1)\equiv 0 \pmod 5 $$

    And hence the solutions to each equation are $x=\pm 1$. Now, we break that into linear systems in the following manner:

    $$x+1\equiv 0 \pmod 3 \qquad x-1\equiv 0 \pmod 3 \ x+1\equiv 0 \pmod 5\qquad x+1\equiv 0 \pmod 5 $$

    $$x+1\equiv 0 \pmod 3 \qquad x-1\equiv 0 \pmod 3 \ x-1\equiv 0 \pmod 5\qquad x-1\equiv 0 \pmod 5$$

     – Red Banana Dec 09 '20 at 08:46
  • 1
    @QiaochuYuan And then, the solutions we get from all these systems are the solutions of the original system and here is where we can actually use the CRT. Is that it? – Red Banana Dec 09 '20 at 08:46
  • 1
    Yes, exactly right! – Qiaochu Yuan Dec 09 '20 at 08:58

2 Answers2

2

let $p$ be a prime number

$x^2\equiv 1 \pmod p$ $\Rightarrow$ $x^2-1\equiv (x-1)(x+1)\equiv 0 \pmod p$

$(x-1)(x+1)\equiv 0 \pmod p$ $\Rightarrow$ $x \equiv 1$ or $x \equiv -1$

Let $n=p_1p_2...p_n$ and $x^2\equiv 1 \pmod n$

$x^2 \equiv 1 \pmod {p_1} $ $\Rightarrow$ $x \equiv \pm1 \pmod{p_1} $

$x^2 \equiv 1 \pmod{p_2}$ $\Rightarrow$ $x \equiv \pm1 \pmod{p_2} $

$\vdots$

$x^2 \equiv 1 \pmod{p_n}$ $\Rightarrow$ $x \equiv \pm1 \pmod{p_n} $

we know that for $a_i \in \{-1,1\}$ $(a_1,a_2,...,a_n)=(x \pmod{p_1},x \pmod{p_2},...,x \pmod{p_n})$ shows a solution for $x^2 \equiv 1 \pmod{n}$

there are $n$ primes and $2$ options for per prime so there are $2^n $ solutions and for all $m \in N^+$ there exists a $n \in N$ such that $2^{n}<m<2^{n+1}$ which proves your statement

Nevzat Eren Akkaya
  • 1,564
  • 1
    According to CRT, there exists a $x$ such that $x=p_1p_2...p_n+c$ and $x \equiv a_1 \pmod{p_1}$ $\cdots$ $x \equiv a_n \pmod{p_n}$ but also it is a fact that $x^2 \equiv 1 \pmod{n}$ let $a_1=1$ if you change $a_1=-1$ then you will obtain another solution for $x=p_1p_2...p_n+k$ where $k \neq c$ but $x^2=(p_1p_2...p_n+k)^2 \equiv 1 \pmod{n}$ so we obtain a different solution for $x^2 \equiv 1$ you can apply this to all $p_i$ and obtain different solutions so $(a_1,a_2,...,a_n)$ shows different solutions for different $a_i$ – Nevzat Eren Akkaya Dec 09 '20 at 08:32
0

For this for any m we need to have:

$x=\underbrace{p_1p_2p_3p_4 \cdot\cdot\cdot}_{m\space times} ±1$

Therefore:

$x^2=P(t= \underbrace{p_1p_2p_3p_4 \cdot\cdot\cdot}_{m\space times}) +1$

Where P(t)represent a polynmial having factors $p_1p_2p_3...$.

In this case any prime factor $(p_i)$ can be considered as n such that :

$x^2\equiv 1 \mod (n)$

sirous
  • 10,751