Integer combinations of two irrational numbers equal to an integer Let $x,y$ be two irrational numbers in $(0,1)$ with the property that for some non-zero integers $a,b,c$ we have $ax+by=c$. I have a general question and a more specific related question. 
The general question is: what can we say about the set of all integers $c$ that can be written this way as an integer combination of $x$ and $y$? In particular must there be many such $c$ if we know that there is in particular one non-zero such $c$, or can it happen (for some $x,y$) that such a $c$ is unique?
The more specific question is: if there is one such combination $ax+by=c$, must there also exist a combination $a'x+b'y=c'$ where $a',b',c'$ are still non-zero integers and in addition $a',b'$ are coprime?
 A: Any given integer $c$ can be represented that way. Take $x=\frac{\pi}{4}=0.785...$, $y=1-\frac{\pi}{4}=0.214...$, $a=b=c$, then $ax+by=c$.
For fixed $x,y\in(0,1)$ irrational let $a,b,c$ non zero integers such that $ax+by=c$. For any integer k you then have $(ka)x+(kb)y=kc$, so the $a,b,c$ are not unique. 
Assume for certain irrational $x,y$ we have $2x+2y=1$, i.e. $x+y=\frac{1}{2}$ (take e.g. $x=\frac{\pi}{8},y=\frac{1}{2}-x$, then $x,y\in(0,1)$). We need to show that for any coprime non zero $a,b$ that $ax+by$ is no integer. Assume there would be such a non zero integer $c$ with $ax+by=c$, then $y=\frac{c-ax}{b}$, plugging that into our formula above and multiplying by $b$ gives $$b=2xb+2c-2ax=2x(b-a)+2c.$$ $a=b=1$ cannot holds since that would imply $c = \frac{1}{2}$. Similary $a=b=-1$ would imply $c=-\frac{1}{2}$. $-1\neq a=b\neq 1$ cannot be since $a$ and $b$ are coprime. So we have $b-a\neq 0$, hence $$x=\frac{b-2c}{2(b-a)}$$ which is a contradiction to $x$ being irrational. So there exists no $a,b,c$ non zero integer with $ax+by=c$.
