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Timeline for How many $n$-adic solutions of $y^{2 \cdot n-1} = y$ are there in $Z_n$?

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2 days ago comment added Marco Ripà Thank you. Edited.
2 days ago history edited Marco Ripà CC BY-SA 4.0
Edited for clarity of notation ($\mathbb{Z}_n$ as the $n$-adic integers) and to correct a couple of $a(n)$ values for $n>13$.
2 days ago history edited LSpice CC BY-SA 4.0
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2 days ago comment added pedroelpanda @MarcoRipà A field is a commutative ring with identity, in which every element has a multiplicative inverse (for each $a$, there is an $a^{-1}$ with $aa^{-1}=a^{-1}a=1$). An integral domain is a commutative ring without zero divisors ($ab=0$ means that either $a=0$ or $b=0$). Since you insist that by $\mathbb{Z}_n$ you mean the n-adic integers, then this is never a field, but it is an integral domain when n is prime.
2 days ago comment added Gerry Myerson I hope that in your submission to oeis you have done a better job of explaining your use of notation, in particular your use of $Z_n$. And I don't think you have engaged with the point raised here that the $n$-adic ring $Z_n$ is never a field, not even when $n$ is prime.
2 days ago history edited Marco Ripà CC BY-SA 4.0
Adding a P.S. which answers to the main question (thanks to Chris' explanationatory comments below).
2 days ago comment added Marco Ripà Thank you (I got it), this settles the general case. I had already proved the part $a(p^m)=a(p)$ for each prime $p$ (using the structure of $(\mathbb{Z}/p^m\mathbb{Z})^\times$), but I had not extended the argument to mixed $n$. Now your $n$-adic CRT decomposition makes the multiplicativity of $a(n)$ immediate (even to this self-taught independent researcher who is writing the present comment).
2 days ago comment added Chris Wuthrich As I said, $a$ is multiplicative and for odd primes $p$, we have $a(p^m) = a(p) = p$ since there are $p-1$ solutions in $\mathbb{Z}_p$ to $y^{2(p-1)}=1$ and one solution to $y=0$. Finally $a(2^m) = 3$.
2 days ago comment added Marco Ripà I used CRT to compute the solutions I gave, but here I'm asking for a general formula to do so easily in order to generate $a(n)$ (for the OEIS I need a(n) up to $n=80$ or so, while for my encryption algorithm I'd like to reach $5 \cdot 10^6$, if possible).
2 days ago comment added Chris Wuthrich This is just the Chinese reminder theorem before taking limits. Not really research level maths, I am afraid.
2 days ago comment added Marco Ripà I know (as I stated in my original post) that if $n := p^m$ is a prime power, than $a(n) = a(p) = p$, but I cannot understand how to compute (at least the number of) solutions of $y^{2 \cdot n-1}=y$ as $n$ is, say, a non-squarefree and non-prime-power integer (such as $756 = 2^2 \cdot 3^3 \cdot 7$ or so).
2 days ago comment added Chris Wuthrich If the prime divisors of $n$ are $p_1$, ..., $p_r$, then $\varprojlim \mathbb{Z}/n^k\mathbb{Z}$ is isomorphic to $\mathbb{Z}_{p_1} \times \cdots\times\mathbb{Z}_{p_r}$. Therefore your $a$ is multiplicative and $a(p^m) = a(p)$. Now this reduces to solving $y^{2(p-1)} - 1=0$ in $\mathbb{Z}_p$ for primes $p$, which is easy.
2 days ago comment added Marco Ripà E.g., for $n=6$ we need to include all (and only) the following solutions: $\{4^{3^n}\}_\infty - \{3^{2^n}\}_\infty, -\{4^{3^n}\}_\infty, \{3^{2^n}\}_\infty, -\{3^{2^n}\}_\infty, \{4^{3^n}\}_\infty, \{3^{2^n}\}_\infty -\{4^{3^n}\}_\infty, -1_6, 1_6, 0_6 $ (where I'm using a compact notation to indicate the inverse limit).
2 days ago comment added Marco Ripà Just to clarify: in this question $\mathbb{Z}_n$ denotes the ring of $n$-adic integers (as in my previous paper on the congruence speed formula), not the finite ring $\frac{\mathbb{Z}}{n\mathbb{Z}}$. The term “$n$-adic” was only meant by analogy in the draft of the question. Here a solution of $y^{2 \cdot n-1}=y$ means $y^{2 \cdot n-1} \equiv y \pmod{n^k}$ for all $k \geq 1$.
2 days ago comment added LSpice Re, OK. I also edited my comment—apparently as you were responding to it; sorry!—to point out that, as @‍ChrisWuthrich and @‍HenriCohen mention, you have not addressed whether $\mathbb Z_n$ indicates a finite ring (that is sometimes a field) or the expected projective limit (that is never a field). It seems that the latter must be what you mean; perhaps you meant just "integral domain"?
2 days ago comment added Marco Ripà @LSpice I know... I submitted it one week ago and it is still under review/on hold (it is A390915).
2 days ago comment added LSpice Your sequence is not known to OEIS. In your comment, you still do not address: either $\mathbb Z_n$ means $\mathbb Z/n\mathbb Z$, in which case it is a field exactly for $n$ prime; or it means $\varprojlim\limits_{k \to \infty} \mathbb Z/n^k\mathbb Z$, in which case it is never a field for $n > 1$.
2 days ago comment added Marco Ripà The equation $y^3=y$ (and thus also $y^5=y$, $y^7=y$, and so on) in $\mathbb{Z}_6$ has exactly $9$ solutions (including $0$, $1_6=\ldots 00001$ and $-1_6=\ldots 5555$) since $6$ is not a prime number and thus $\mathbb{Z}_6$ is a mere commutative ring and not a field (as $\mathbb{Z}_3$, $\mathbb{Z}_5$, and so forth). If you are interested in the list of the mentioned solutions, just look at Equation (3.5) of our recent preprint: zenodo.org/records/17744007.
2 days ago history edited LSpice CC BY-SA 4.0
List of values of $a(n)$ as a table
2 days ago comment added Henri Cohen Indeed confusing, Chris is right, but since for prime n Z_n is a field, the second interpretation is correct, but then Z_6 has 6 elements, how is it possible that you find 9 solutions ?
2 days ago history edited GH from MO
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2 days ago comment added Chris Wuthrich You talk about $n$-adic, but towards the end it looks like $\mathbb{Z}_n$ does not stand for the ring of $n$-adics, but for $\mathbb{Z}/n\mathbb{Z}$. Very confusing.
2 days ago history edited Marco Ripà CC BY-SA 4.0
minor edit
2 days ago history asked Marco Ripà CC BY-SA 4.0
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