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Is it possible to show (or when is it true); that for the group $\mathbb{Z}^+_n :=(\mathbb{Z}_n , +_n) $, there exists an $a \in \mathbb{Z}^+_n$ for each $z \in \mathbb{Z}^+_n$, where both $z+_n a$ and $z+_n -a$ are totatives of $n$?

Additional What is the smallest $a$ that can be constructed?

Brad Graham
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  • What does it mean to be totative? – Tobias Kildetoft Feb 11 '16 at 11:32
  • A totative of some number $n :=$ a number that is coprime to $n$. – Brad Graham Feb 11 '16 at 11:35
  • I guess it is true. If $hcf(z,n)=z$ then we can pick an $a$ which is a product of all prime divisors of $n$ except those prime numbers which are factors of $z$. In that case both $z+a$ and $z-a$ wont be divisible by any factor of $z$. – Brad Graham Feb 11 '16 at 11:46
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    Actually, you do not seem to even be using that $z$ divides $n$ there. Just pick $a$ as the product of those primes which divide $n$ but $z$. – Tobias Kildetoft Feb 11 '16 at 11:51
  • Considering $n=210$ and $z=8$. And naturally choosing $a=105$. It's true that $113$ (repeated) is a totative of $210$. But it is the case that if we choose $a=21$, both $29$ and $197$ are also totatives of $210$. So I guess my question is, why is that? – Brad Graham Feb 11 '16 at 11:59
  • You mean why we can pick more than one $a$ that works? That is just because there is a pretty decent number of totatives (relatively). – Tobias Kildetoft Feb 11 '16 at 12:01
  • My aim is to show that there is a small enough $a$. Taking $n$ to be a primorial number $ P_x# := p_1 \times p_2 \times ... \times p_x$ ($p$ are primes), I am wondering if there exists an $a< \frac{1}{2} p_x^2$. – Brad Graham Feb 11 '16 at 12:07
  • Do you want to say "for each $z \in \mathbb{Z}^+_n$ there exists an $a \in \mathbb{Z}^+_n$..."? – user26857 Feb 11 '16 at 13:32
  • @user26857 Exactly, the passive expression is equivalent in meaning, only the focus differs. From my interest, the focus on the existence of $a$ occurred naturally due to my comment before. – Brad Graham Feb 11 '16 at 13:53
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    Hint: see this answer. – Bill Dubuque Jul 17 '16 at 23:37

1 Answers1

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If you want to show that

for each $z\in\mathbb{Z}_n$ there exists $a\in\mathbb{Z}_n$ such that $\gcd(z\pm a,n)=1$,

then proceed like this:
If $\gcd(z,n)=1$ then set $a=0$.
Otherwise $\gcd(z,n)>1$. If for any prime $p$ such that $p\mid n$ we have $p\mid z$ then set $a=1$. Otherwise, there is $p\mid n$ such that $p\nmid z$. In this case let $a$ be the product of all primes in $n$ which don't appear in $z$.

user26857
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  • Thank you. So for additional points, is it possible to construct a suitable $a$, which is smaller than the product of all primes in $n$ which don't appear in $z$? (Consider $n=210$, $z=8$. Your method sets $a=105$, which obeys the criteria. But so does $a=21$). – Brad Graham Feb 11 '16 at 15:12
  • Careful, $5$ is not a totative of $210$ – Brad Graham Feb 11 '16 at 15:17
  • Actually, I can explain the case mentioned above, by considering a $8$ modulo $2,3,5$ and $7$; $21$ modulo $2,3,5$ and $7$, and recognizing that totatives of $210$ are not equivalent to $0$ modulo $2,3,5$ or $7$. So by sort of cross analysis of modulos we can find all $a$ for a $z$ I think... – Brad Graham Feb 11 '16 at 15:29
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    Maybe finding a minimal $a$ is not an easy task. See here. – user26857 Feb 11 '16 at 15:30
  • You could also think on this problem by using CRT. Write $n=p_1^{k_1}\cdots p_r^{k_r}$. Then $z$ corresponds to some $(z_1,\dots,z_r)$ and $a$ to an $(a_1,\dots,a_r)$, where $z_i,a_i\in\mathbb Z_{p_i^{k_i}}$. You need $z_i-a_i$ invertible in $\mathbb Z_{p_i^{k_i}}$. This way is very easy to find a suitable $a_i$, but the minimality is a different story. It seems to be difficult even for $n$ square-free. – user26857 Feb 11 '16 at 15:36
  • Thanks alot for the link. If we can show that for $n$ being a primorial number $P_x# := p_1 \times p_2 \times ... \times p_x$ (where $p$ are consecutive primes), that for all $z$ there exists an $a <\frac{1}{2}p^2_x$, then we have proved Goldbach's conjecture. So I would be disappointed if it wasn't difficult! – Brad Graham Feb 11 '16 at 15:42