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I claimed in a recent post that

Let $p_1=2,p_2=3,p_3,...$ the prime numbers. $$\forall n \in \mathbb N^*,p_n\#:=p_1...p_n\text{ (n-th primorial)}$$ $$U_{p_n\#}:=(\mathbb Z/p_n\#\mathbb Z)^\times$$ $$J_n:=\{u\in U_{p_n\#}|u+2\in U_{p_n\#}\}$$ $$\boxed{\text{result :}|J_n|=(p_n-2)...(p_2-2),\text{ where }|J_n| \text{ is the number of elements of }J_n}$$

It should be remembered that

  • for $m\in \mathbb N$,$(\mathbb Z/m\mathbb Z)^\times:=\{u\in \mathbb Z/m\mathbb Z: \exists v\in Z/m\mathbb Z, uv=1\}$
  • thanks to Bézout's theorem, we then have a simple description of these units.

As no one has offered me any references for the moment, I am reduced to demonstrating the result claimed by myself. Here's what I propose : by induction , cases $$\mathbb Z/6\mathbb Z(n=2, J_2=\{-1\}, |J_2|=1), \mathbb Z/30\mathbb Z(n=3, J_3=\{11, 17,-1\}, |J_3|=3=(5-2)(3-2))$$being easy to check directly.

I know that $\mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z \times \dots\times \mathbb Z/p_{n+1}\mathbb Z$ is isomorphic to $\mathbb Z/p_{n+1}\#\mathbb Z$(isomorphism of rings)

$$u=(x_1,...,x_n,x_{n+1})\in J_{n+1}\iff \begin{cases} x_{n+1} \neq 0 \\ x_{n+1}+2\neq 0\\ \forall i\in [1,n] x_i+2 \neq 0 \end{cases}$$ $$\iff (x_1,...,x_n)\in J_n\land x_{n+1}\neq 0,-2$$

Then $|J_{n+1}|=(p_{n+1}-2)\times |J_n|$

And the claimed result is thus obtained by induction.

I don't see any major problem with my proof. For me, the difficulty is, I repeat, that I have no reference and have not seen this proof anywhere, not even the introduction of the object $J_n$. I am therefore very interested in your opinion on the validity of the proof as well as your superior knowledge of the subject.

Stéphane Jaouen
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  • Presumably, if your definition of $J_n,$ you aren't assuming $p$ is prime. It is bad to use $p$ to mean an arbitrary integer in a problem which involves primes - better to us $m,$ $k,$ $j.$ Also, often $U_m$ is defined to be $(\mathbb Z/n)^{\times},$ which might confuse your reader here. – Thomas Andrews Mar 30 '24 at 16:53
  • @ThomasAndrews : Thanks, I edited, hope it's better and responds to your criticisms – Stéphane Jaouen Mar 30 '24 at 17:00
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    You should also define the group of units $U_n=(\Bbb Z/n\Bbb Z)^{\times}$ in general. Note that $\Bbb Z/n\Bbb Z$ is a ring here (not an algebra). – Dietrich Burde Mar 30 '24 at 17:04
  • For me, an algebra is a $K$-vector space with a bilinear product. – Dietrich Burde Mar 30 '24 at 19:49
  • Yes, it is a vocabulary. But usually $\Bbb Z/n$ is considered as a ring, and its units are denoted by $U(n)$ or $U_n$. – Dietrich Burde Mar 30 '24 at 19:51

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