Existence of $a,b$ relatively prime to $2k$ such that $a + b \equiv 2 \pmod{2k}$ Let $k$ be a positive integer. Then, I would like to know when there are $a,b \in (\mathbb{Z}/(2k))^{\times}$ with $a \neq b$ so that $a + b \equiv 2\pmod{2k}$.
I have computational evidence to suggest that this should be true whenever $k \neq 1,2,3,6$. However, I am having a bit of trouble proving this. Here are some things that I have thought about so far:

*

*First, I approached this problem using a "sieving" technique on $2k$. For instance, $(a,b) = (3,-1)$ is always a solution whenever $2k$ is not divisible by $3$, so we need to worry about only the cases where $2k$ is divisible by $3$. Using similar logic, we can show that any pair of twin primes $(p, p+2)$ would give a solution $(a,b) = (p+2, -p)$ if $2k$ is not divisible by $p$ and $p+2$, i.e., we need to only consider cases where $2k$ is divisible by at least one prime from each pair of twin primes (of course, there are some constraints on making sure $a \neq b$, but I've brushed those under the rug for the time being). Then, if the twin prime conjecture were true, we'd see that such an integer $k$ could not exist...but of course we don't know that :P Some alternative techniques that I've tried here is to consider primes of the form $p^m \pm 2$, where $p$ is a prime and $m$ some positive integer, but I wasn't able to find out much about the distribution of such primes.


*It should be (roughly) equivalent to show that there exists some $a \in (\mathbb{Z}/(2k))^{\times}$ such that $a - 2 \in (\mathbb{Z}/(2k))^{\times}$ as well. This is because if $a - 2 \in (\mathbb{Z}/(2k))^{\times}$, then $2-a \in (\mathbb{Z}/(2k))^{\times}$ as well. Conversely, if $a + b \equiv 2 \pmod{2k}$ with $a,b$ both relatively prime to $2k$, then $b = 2 - a$ and so $a-2$ is relatively prime to $k$ as well. The "roughly" is because we'll need to impose that $a \not\equiv 2 - a \pmod{2k}.$


*Using (2), I tried attacking this problem using a pigeonhole-esque strategy: if such an $a$ didn't exist in $(\mathbb{Z}/(2k))^{\times}$, then the "gap" between elements in $(\mathbb{Z}/(2k))^{\times}$ is at least $3$ (the gap can't be equal to $1$ because $2k$ is even). Then, this roughly means that there are at most $2k/3$ elements in $(\mathbb{Z}/(2k))^{\times}$, which rules out some choices for $k$. However, since $\phi(2k)$ (the Euler totient function) has a sublinear lower bound, it seems that we are still left with infinitely many integers.
In light of the above, I was wondering whether I could have some advice on what to think about next.
 A: You solved the case $3\nmid k$.
Suppose $k=9m$. Then $a=1+6m$, $b=1-6m$ is a solution.
Remains the case $k=3u$ with $3\nmid u$. You already know the exceptions $k=3$, $k=6$, hence we may assume $u>2$.
Use the Chinese Remainder Theorem to solve
$$ a\equiv 1\pmod {3},\quad a\equiv 3\pmod {2u} $$
for $a$ as well as
$$ b\equiv 1\pmod {3},\quad b\equiv -1\pmod {2u} $$
for $b$.
Then

*

*$a,b$ are coprime to $2k$,

*$a+b\equiv 2$ both $\pmod{3}$ and $\pmod {2u}$, hence also $\pmod{2k}$.

*$a\ne b$ becasue $u>2$ implies $3\not\equiv-1\pmod {2u}$
A: a and b are both odd numbers. Let $a=b+2t$ we can write:
$$a+b=2b+2t\equiv2\bmod 2k\Rightarrow b+t=1+mk$$
where $m=1, 2, 3, . . .$
this means we can always construct a relation such as $a+b\equiv 2\bmod 2k$ if $b+t\equiv 1 \bmod k$
For example $b=3, t=2\rightarrow a=3+4=7$ and we have to have $k=b+t-1$,  we have $k=5$ and:
$$a+b=7+5=12\equiv 2\bmod 2\times 5$$
Or:
$b=14, t=4\rightarrow a=22$
$k=14+4-1\rightarrow k=17$
and we have:
$a+b=14+22=36=2+2\times17\equiv 2\bmod 2\times 17$
