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I am trying to solve the following problem:

enter image description here

Given the context in the book, I have noticed the following: Suppose $n=30$ then we write the following system of equations:

$$x^2\equiv 1 \text{ (mod 2)}\\x^2\equiv 1 \text{ (mod 3)}\\x^2\equiv 1 \text{ (mod 5)}$$

And use the chinese remainder theorem. The solutions for $x^2\equiv 1 \text{ (mod 30)}$ are then the solutions for the previous system. I made a table on Mathematica which seems to confirm my suspicion. Whenever the previous system of equations is satisfied, $x^2\equiv 1 \text{ (mod 30)}$.

$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

I have two questions:

  • Why the solutions of that equation are found in that system of equations?
  • How can I count the solutions? I kinda guessed the previous one and it seems to work but I have no clue on how to count them. I've been able to see that there must be an even number of solutions because $a^2\equiv (n-a)^2 \text{ (mod n)}$ but aside from this, I didn't make much progress.
Red Banana
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1 Answers1

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The Chinese remainder theorem, in its abstract version, asserts that the map \begin{align} \mathbf Z/30\mathbf Z&\longrightarrow \mathbf Z/2\mathbf Z\times \mathbf Z/3\mathbf Z\times \mathbf Z/5\mathbf Z\\ n\bmod 30&\longmapsto(n\bmod 2,n\bmod3,n\bmod 5) \end{align} is a ring isomorphism.

Therefore, to count the solutions, you count them modulo $2$ (1), modulo $3$ and modulo $5$ (two each, since the quotients are fields) and combine them in all possible ways, which makes $4$ solutions modulo $30$ in all.

J. W. Tanner
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Bernard
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  • But is it possible to obtain a closed formula for the solutions? – Red Banana Jun 15 '20 at 20:25
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    I'd expect that for a general formula, the number of powers of 2 will be a special case: if I'm not mistaken, for an odd prime $p$, there would be exactly two solutions to $x^2\equiv 1 \pmod{p^k}$ for $k \ge 1$, whereas there are exactly two solutions to $x^2 \equiv 1 \pmod{2}$, four solutions to $x^2 \equiv 1 \pmod{4}$, and eight solutions to $x^2 \equiv 1 \pmod{2^k}$ for $k \ge 3$. (The proofs of these facts would most likely involve what's known as "Hensel's lemma".) – Daniel Schepler Jun 15 '20 at 20:36
  • @BillyRubina: A closed formula for this particular $n$ or a closed formula for any $n$? – Bernard Jun 15 '20 at 20:50
  • @Bernard For this case. The next exercise asks me to show that for any $m$, there exists $n$ such that the equation $x^2=1$ has at least $m$ solutions in $\mathbb{Z}/(n)$. Perhaps it's not needed to have a closed formula. – Red Banana Jun 15 '20 at 21:27
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    There exists a simple formula, based on a Bézout's relation, for the inverse isomorphism in the case of two factors, which may be used repeatedly if there are morer factors. For $n=30$, you obtain that the four solutions are $x\equiv \pm 1, \pm11\mod 30$. – Bernard Jun 15 '20 at 21:55