Suppose $a,b\in\Bbb N$ are odd coprime with $a,b>1$ then is it true that if all four of $$x\_1a+x\_2b,\mbox{ }x\_2a-x\_1b,\mbox{ }x\_1\frac{(a+b)}2+x\_2\frac{(a-b)}2,\mbox{ }x\_2\frac{(a+b)}2-x\_1\frac{(a-b)}2$$ are in $\Bbb Z$ for some $x\_1,x\_2\in\Bbb R$ then $x\_1,x\_2\in\Bbb Z$ should hold?



A simple number theory confirmation
Yes. If we set $\alpha = (a+b)/2$ and $\beta=(a-b)/2$, then the lattice generated by the four vectors


$$
\binom{\alpha}{\beta},\binom{-\beta}{\alpha},\binom{\alpha+\beta}{\alpha -\beta},\binom{\beta -\alpha}{\alpha +\beta}
$$
is contained in the set of $(y\_1,y\_2) \in \mathbb{Z}^2$ such that $x\_1 y\_1 + x\_2 y\_2 \in \mathbb{Z}$. 


We thus have to check that these four vectors generate all of $\mathbb{Z}^2$. This is true whenever $\alpha$ and $\beta$ are coprime integers with $\alpha + \beta$ odd. Indeed, these conditions imply that the three integers $\alpha^2 + \beta^2$, $\alpha^2 - 2\alpha \beta - \beta^2$, $\alpha^2 + 2\alpha \beta - \beta^2$ are coprime; this ensures the existence of three integers $u,v,w$ such that the quantity
$$
\mathrm{det} \left( \binom{\alpha}{\beta},u \binom{-\beta}{\alpha}+v\binom{\alpha+\beta}{\alpha -\beta}+w\binom{\beta -\alpha}{\alpha +\beta} \right) \\= u(\alpha^2 + \beta^2) + v(\alpha^2 - 2\alpha \beta - \beta^2)+w(\alpha^2 + 2\alpha \beta - \beta^2)
$$
is equal to $1$ (Bezout's lemma). This shows that the two vectors
$$
\binom{\alpha}{\beta}, \ \ \mathrm{and} \ \ u \binom{-\beta}{\alpha}+v\binom{\alpha+\beta}{\alpha -\beta}+w\binom{\beta -\alpha}{\alpha +\beta}
$$
form an integral basis of $\mathbb{Z}^2$.


