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Given coprime $a,b\in[2^{n+k},2^{n+k+1})\cap\Bbb Z$ with $|a-b|>2^{n+k-1}$ and given $m\in[2^{2n+k},2^{2n+k+1})\cap\Bbb Z$ consider the equation

$$ax+by+z=m$$ where $x,y,z\in\Bbb Z_{>0}$ are unknown.

Due to size bounds on $a,b,m$ and because $x,y,z>0$ holds we cannot have $x,y>0$ for every $z>0$ if $k>n$.

If $k>n$ how often is there is an unique $z>0$ such that $x,y>0$ holds?

For any $k>n$ at most how many $z>0$ do we have $x,y>0$?

I ask for what integers $z>0$ we have $ax+by+z=m$ with $x,y>0$. I think there is only one most of the time and always one all the time. So I am seeking to quantify 'most'. It depends on minimum solutions to $ax'+by'=1$. If $|x'|,|y'|$ are large enough there should be only one exactly. However there could be many depending on size of $|x'|,|y'|$. How many $z>0$ are there?

WLOG let $x'>0$ and $y'<0$ in $ax'+by'=1$ so that $x'+|y'|$ is least.

Then if $x_0,y_0$ is solution at $z_0>0$ with $ax_0+by_0+z_0=m$ then we have for all small $c\in\Bbb N$ the form $ax_0+by_0+cax'+cby'+z_0-c=m$ and so $ax''+by''+z''=m$ holds where $x''=x_0+cx'>0$ holds. However $y''=y_0-c|y'|$ could be less than zero. This will definitely happen as soon as $c$ exceeds a threshold. Then my query is how often is $c>0$ for this to happen?

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    Please clarify this question. What do you mean by "$x,y>0$ for every $z>0$"? I thought you were already requiring $x,y>0$. Do you mean to ask for what integers $z>0$ there is an integer solution of $ax+by+z=m$ with $x,y>0$? – 2017-01-12
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    @RobertIsrael yes I ask for what integers $z>0$ we have $ax+by+z=m$ with $x,y>0$. I think there is only one most of the time and always one all the time. So I am seeking to quantify most. It depends on minimum solutions to $ax'+by'=1$. If $|x'|,|y'|$ are large enough there should be only one exactly. However there could be many depending on size of $|x'|,|y'|$. How many $z>0$ are there? – 2017-01-12

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WLOG let's suppose $a < b$. Since $z = m - ax - by$, the greatest possible $z$ corresponds to the least possible $x$ and $y$: $x=y=1$, thus $z = m - a - b$. Notice that $m - a - b \ge 2^{2n} - 7 \cdot 2^{n-1} > 0$ if $n \ge 2$. The next greatest would be $z = m - 2a - b$ for $x=2$, $y=1$, and $m - 2a - b \ge 2^{2n} - 5 \cdot 2^n > 0$ if $n \ge 3$.

In general, we find that for any $M$ there will be some $N$ such that if $n \ge N$ there are at least $M$ different possible $z$.

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    I want something on 'at most' some number of solutions. Also what is $M$ and $N$? and $m,a,b$ are not arbitrary they are in specified intervals. – 2017-01-12
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    I agree with your post. However I forgot something. I need $k>n$. – 2017-01-12