(Original title: Whether certain subgroups of $\mathbb{Z}^2$ contain $(1, 0)$)
Consider subgroups of $(\mathbb{Z}^2, +)$ of the form $H = \left\langle (a, b),\ (b, a),\ (-a, b),\ (-b, a) \right\rangle$ where $a, b \in \mathbb{Z}^+$ and $a < b$. I want to find out for which conditions on $a, b$, $H$ contains $(1, 0)$. I find it interesting because when $(1, 0) \in H$ it's easy to see that $v \in H \ \ \forall v \in \mathbb{Z}^2$.
This is equivalent to asking whether we can find $\lambda,\ \lambda',\ \mu,\ \mu' \in \mathbb{Z}$ such that $\lambda(a, b) + \lambda'(b, a) + \mu(-a, b) + \mu'(-b, a) = (1, 0)$ thus solving the linear system $$(*)\begin{cases} \lambda a + \lambda' b - \mu a - \mu' b = 1\\ \lambda b + \lambda' a + \mu b + \mu' a = 0\end{cases}$$ I've shown that when $a, b$ are coprime they must be of different parity. Here's a quick draft:
Suppose $a, b$ coprime, then from the second equation of $(*)$ we have $\lambda' + \mu' = hb$ and $\lambda + \mu = -ha$ for some $h \in \mathbb{Z}$. From this we can find $\lambda$ and $\lambda'$ and substitute them into the first equation, finding $h(b^2 - a^2) = 1 + 2(a\mu + b\mu')$. Since both sides of the equation are integers they must have the same parity, so $h(b^2 - a^2)$ must be odd and that happens when $h$ is odd and $a, b$ have different parity.
From here on I'm kind of stumped.
Is this line of reasoning correct if I wanted to prove that when $a, b$ satisfy the previous conditions then a solution to $(*)$ does in fact exist? Also would it be possible to find explicitly one of such solutions for generic $a, b$?
Here are two plots of possible solutions for $(a, b) = (1, 10)$ and $(a, b) = (3, 8)$
Edit
After working on the problem a bit more I think I can rephrase the question this way.
Edit 2
I made a wrong, stupid assumption. I think the question is now correctly formulated.
By letting $h = 1$ we can find $\mu'$ in terms of $\mu$, namely $\mu’ = \frac{b^2 - a^2 - 1 - 2a\mu}{2b}$.Now, obviously we need both variable to be integers so now the problem is to find an adequate value for $\mu$.
My candidate would be $\mu = \frac{b^2 - a^2 - 1 - 2b}{2a}$, if you assume that it is an integer and substitute it into the previous equations it should give a correct generic solution for the problem.
So, right now the real question would be: For which $a, b, \mu$ is $\frac{b^2 - a^2 - 1 - 2a\mu}{2b}$ an integer?. (and I'm changing the title of the question to make that clear)
According to the previous part of my question, it can be seen that if $b = ak,\ k \in \mathbb{Z}$, the resulting fraction $\frac{a(a(k^2-1) - 2\mu) - 1}{2ak}$. That fraction can never be an integer as the numerator is never a multiple of $a$, so it still holds the condition that $a, b$ must be coprime.
I don't understand why all this is necessary? We write the equation.
$$\frac{b^2-a^2-1-2b}{2a}=q$$
Decompose into factors the number so that the solutions were integers. $q^2-2=tk$
And solutions record.
$$b=\frac{k-t}{2}+1$$
$$a=-q\pm\frac{k+t}{2}$$
The problem is reduced to the decomposition of numbers difference of squares.
Another equation? Well!
$$\frac{b^2-a^2-1-2ja}{2b}=q$$
Use. $j^2-q^2-1=tk$
$$b=\frac{k-t}{2}+q$$
$$a=-j\pm\frac{k+t}{2}$$