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I am stuck with this exercise:

Show that $\forall s >0 \quad \exists N >0$ such that $$x^2 \equiv 1\pmod N$$ has more than $s$ solutions.

I think I cannot use the theory if quadratic residues, because the modulus is not prime, so maybe I must prove it with square roots or roots of unity, but I don't know how to approach it at all.

Thanks for any help.

Phi_24
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2 Answers2

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Hint:

Consider a square-free integer: $\;N=p_1p_2\dotsm p_r$. By the Chinese Remainder theorem, solving the equation $x^2\equiv 1\pmod N$ is equivalent to solving the set of congruences $$\begin{cases} x^2\equiv 1\pmod{p_1} \\ \quad\enspace\vdots\\x^2\equiv 1\pmod{p_r} \end{cases}\iff \begin{cases} x\equiv\pm 1\pmod{p_1} \\ \quad\,\vdots\\x \equiv \pm1\pmod{p_r} \end{cases}$$ so there are $2^{r-1}$ solutions.

Edit. Since $2\mid N$ for any $r \geq 1$, the answer for $N = p_r\#$ is $2^{r-1}$ as there is only one solution modulo $2$, while there are two solutions modulo any other prime.

Daniel Donnelly
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Bernard
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    Is not $2^r$ the number of solutions? – ajotatxe Jun 19 '19 at 15:01
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    Oops! Yes, of course. Sorry for the mistyping, and thank you for your pointing it! – Bernard Jun 19 '19 at 15:17
  • This was not correct, so I've edited. It is correct if $p_1 = 2 \nmid N$. However, modulo $2$ there is only one solution, as $1 = -1 \mod 2$. Therefore the answer is $2^{\omega(N) - [2\mid N]}$ in general or since $2 \mid N$ in your example, it's $2^{r - 1}$. Where $[2\mid N] = \begin{cases} 1 \text{ if } 2 \mid N \ 0 \text{ if } 2 \nmid N\end{cases}$. Also, the asker asks for general $N$, not necessarily square-free, so what about that case? @Bernard – Daniel Donnelly Feb 23 '25 at 15:48
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Hint $ $ Let $\,f(x) = x^2-1\,$ below and exploit the highlighted $\rm\color{#c00}{multiplicativity}$ of number of roots.

Remark $\ $ If $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ & mod $\,n.\,$ By CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn,\,$ i.e.

$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{mn}&\overset{\rm \large CRT\!\!}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_{\color{}{\large k}}\pmod{\! m},\phantom{I^{I^{I^I}}}\ \ \color{#c00}k\ \ \rm roots\\x\equiv s_1,\ldots,s_{\large\color{}{\ell}}\pmod n,\ \ \ \ \ \ \ \ \ \,\color{#c00}{\ell}\ \ \rm roots\end{array}\\[.3em] &\iff& \left\{ \begin{array}{}x\equiv r_{\large i}\pmod{\! m}\\x\equiv s_{\large j}\pmod n\end{array} \right\}\ \ \ \ {\rm for}\ \ \begin{array}{}1 \le i \le k\\ 1\le j\le\ell\end{array}\\ &\overset{\rm\large CRT\!}\iff& \left\{ x\equiv t_{\,\large i\, j}\!\!\!\!\pmod{\!mn} \right\}\,\ \ \underbrace{{\rm for}\ \ 1 \le i\le k,\,\ 1\le j\le\ell}_{\Large \color{#c00}{k\,\cdot\, \ell}\ \,\rm roots\ \ \ \ \ \ \ \ \ \ }\\ \end{eqnarray}\qquad\qquad$$

Bill Dubuque
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    so how do I conclude that the number of solutions is greater than any s? – Phi_24 Jun 19 '19 at 15:12
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    @Phi_24 By above, if we multiply the modulus by a new prime $,p>2,$ we $\rm\color{#c00}{double}$ the number of roots, since $,(x!-!1)(x!+!1),$ has $,\color{#c00}2,$ roots $,\pm1\pmod{!p}.,$ But by Euclid there are infinitely many primes, so we can repeatedly double the number of roots beyond any bound. – Bill Dubuque Jun 19 '19 at 15:34
  • What about for $N$ that is not square-free? I know CRT still applies, but how do we count solutions to $x^2 -1 \mod p^k$ for $k \geq 2$? – Daniel Donnelly Feb 23 '25 at 15:54
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    @Dan The general case is discussed in other answers here. – Bill Dubuque Feb 24 '25 at 00:37