Do the set of coprimes of a number form an additive basis?

Question

For any positive integer $n,$ define $X:= \{ a: (a,n) = 1,\ 1\leq a < n \}.$ Let $(X+X)^*:= \{ (x+y)\pmod n: x,y\in X \}.$

Conjecture: If $n$ is even, then $(X+X)^*= \{ 0,2,4,\ldots,n-2\}.$ If $n$ is odd, then $(X+X)^*= \{ 0,1,2,3,\ldots,n-1\}.$

I do not know if this is known or not, although it seems to be true for small integers $n.$

If so, then perhaps this could somehow link Goldbach's conjecture with Euler's totient function.

One idea of the conjecture is that if we try to omit lots of numbers from $X$ by, for example, choosing $n=2\cdot 3\cdot 5\cdot 7,$ then since $n$ is a large number, all the primes between $7$ and $n$ will be able to "help out" in trying to form the additive basis, making it likely to be an additive basis. Can I prove it continuing down this line of reasoning?

Answer 1

We will prove it by induction.

When $n=p^k$ is a a prime power, then any $2x=x+x$ and $2x=(x-1)+(x+1).$ One of these is a sum of two numbers relatively prime to $n.$

Assume it is true for $n'<n.$ Given $n$ not a prime power, then $n=n_1n_2$ for $n_1,n_2<n$ and $\gcd(n_1,n_2)=1.$ For each $x,$ we can write $$2x\equiv u_i+v_i\pmod {n_i}$$ with $u_i,v_i$ relatively to $n_i.$ Then we can solve the Chinese Remainder Theorem problems:$$u\equiv u_i\pmod{n_i}$$ and $$v\equiv v_i\pmod{n_i}$$ for $i=1,2.$

Show that $u,v$ are both relatively prime to $n=n_1n_2,$ and $2x\equiv u+v\pmod n.$

It is clear that when $n$ is odd, every element can be written as $2x.$ When $n$ is even, it is clear that $u+v$ cannot be an "odd" element.

For example, when $n=60,$ and $2x=4,$ we get $$4\equiv 1+3\pmod{4}\\4\equiv2+2\pmod{3},\\4\equiv 2+2\pmod{5}$$ Giving $u=17,v=47.$

Answer 2

The nontrivial part is to show that in $\:\!\Bbb Z_n\:\!$ every even $\,2a = u+v\,$ is a sum of units (invertibles). More generally it's true if we replace $\:\!2a = a(\color{#90f}{1\!+\!1})$ by $\:\!a\color{#90f}s\:\!$ for $\:\!\color{#90f}{s=u_0+v_0}\:\!$ a sum of units, i.e.

Theorem $ $ In $\,\Bbb Z_n$ the set of sums of two units is closed under scalings $\,s\mapsto a\:\!s\,$ by any $\:\!a\in \Bbb Z_n$.

Proof $ $ (sketch) $ $ We seek unit roots $\,x,y\,$ of the polynomial $\,f(x,y) = x+y-a\:\!s.\,$ As proved here, this has "multiplicative" structure, i.e. if $\:\!n = n_1 n_2\:\!$ for coprime $\:\!n_1,n_2\:\!$ then roots $\!\bmod n\,$ are precisely the CRT lifts of pairs of roots $\!\bmod n_1$ and $\!\bmod n_2.\:\!$ Here we additionally need to invoke: $\:\!u\:\!$ is a unit in $\:\!\Bbb Z_n\:\!$ iff it is a unit in both $\:\!\Bbb Z_{n_i},\:\!$ i.e. in ring language the CRT isomorphism $\,\Bbb Z_n \cong\:\! \Bbb Z_{n_1}\!\times \Bbb Z_{n_2}$ restricts to an isomorphism of units $\,\Bbb U_n\cong\:\! \Bbb U_{n_1}\!\times \Bbb U_{n_2}.\,$ By iterating such decomposition, we completely factor $\:\!n\:\!$ into prime powers $\:\!p^k$, then we can calculate the unit sum for $\:\!as\:\!$ in each component $\:\! \Bbb Z_{p^k},\,$ then apply CRT to lift all of the solutions up to $\:\!\Bbb Z_n,\,$ $ n = \prod p_i^{k_i}.\,$ Finding a solution in $\:\! \Bbb Z_{p^k}\,$ is easy: $\:\!a\:\!$ is a unit iff $\:\!a\:\!$ is coprime to $\:\!p^k$ iff $\,p\nmid a,\,$ so we get two cases

• $\:\!a\:\!$ is a $\rm\color{#0a0}{unit}$ in $\:\!\Bbb Z_{p^k},\,$ thus we have: $\ as = \color{#0a0}{a u_0} + a v_0\,$ is a sum of units.

• $\:\!a\:\!$ $\rm\color{#c00}{nonunit}\!$: $\,p\mid a\:\!$ so $\:\!p\nmid as\!-\!1\:\!$ so $\ as = \color{#c00}1+(as\!-\!1)\,$ is a sum of units. $\ \ \small\bf QED$

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Optimization $ $ When computing solutions we can avoid the above expensive prime factorization. Instead, taking repeated gcds [gdc] with $\,a\,$ we collect into $\,n_2\,$ all the primes in $\,n\,$ that divide $\,a.\,$ These have above $\rm\color{#c00}{nonunit}$ solution $\,as \equiv \color{#c00}1+(as\!-\!1) \pmod{\!n_2}.\:\!$ $\:\!n_1 = n/n_2\,$ is coprime to $\,a\,$ thus here we have the $\rm\color{#0a0}{unit}$ solution $\,as\equiv\, \color{#0a0}{au_0}\,+\,av_0\,\pmod{\!n_1}.\:\!$ Then we compute $\:\!u\!\pmod{\!n}\:\!$ by CRT solving $\,u\equiv \color{#0a0}{au_0}\pmod{\!n_1},\,$ $\,u\equiv \color{#c00}1\pmod{n_2},\,$ then $\,v\equiv as-u\,$ (by $\,as\equiv u+v)$.

Example $\!\bmod 4620\!:\,$ write $\,6(1\!+\!23) \equiv u + v\,$ as a sum of units (via easy mental arithmetic).

$\:\!4620 = 35\cdot 12 = n_1\:\! n_2\,$ by repeatedly moving gcds with $\:\!a\!=\!6\,$ into $\:\!n_2.\:\!$ In each factor $\:\!n_1,n_2\,$ we apply the above solutions, depending on whether or not $\:\!a \!=\! 6\,$ is a unit in $\:\!\Bbb Z_{n_i},\,$ i.e. $\!\bmod n_i$

$\ \,\qquad\qquad\quad\begin{align} \overbrace{6(1\!+\!23)}^{\textstyle 144} &\equiv\,\overbrace{\color{#0a0}{6(1)} + 6(23)}^{\textstyle u_i +\, v_i\,\ }\pmod{\!35}\quad [\:\!6\ \text{is a $\color{#0a0}{\rm unit}$} ]\\[.5em] &\equiv\, \color{#c00}1+(144\!-\!1)\!\pmod{\!12}\quad\:\![\:\!6\:\!\ \color{#c00}{\rm nonunit}]\end{align}$

Applying Easy CRT: $\ \ u\equiv \color{#0a0}6,\color{#c00}1\pmod{\!35,12}$
$\qquad\qquad\quad\ \iff\ u\equiv \color{#0a0}6\!-\!35\left[{\large \frac{\color{#0a0}6\ -\ \color{#c00}1}{35(\equiv\:\! -1)}}\!\bmod 12\right]^{\vphantom{|^|}}\!\!$ $\equiv 6\!-\!35[-5]\equiv \color{#0af}{181}\pmod{\!4620}$

therefore $\bmod 4620\!:\,\ 6(1\!+\!23)\equiv \color{#0af}{181} -37\:\!$ is a sum of units.