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For nonzero integers $x$ and $y$, does there always exist $a$ and $b$ of opposite signs such that $ax+by=\gcd(x,y)$? The property is true for $x,y>0$, due to the gcd being less than or equal to $x$ or $y$, and also for $x,y<0$ since $\gcd(x,y)=\gcd(-x,-y)$. I'm wondering about the cases where either $x$ or $y$ is negative and the other one is positive. I'm aware that countably infinite solutions exist but I don't know if that guarantees $a$ and $b$ of opposite signs. From manually testing some values, I don't believe it is actually possible. Here is what I attempted:

Assume wlog that $x>0$ and $y<0$. Then the solutions $(a,b)=(a_0-k{y\over d},b_0+k{x\over d})$ where $d=\gcd(x,y)$. The solutions only occur for integer $k$ but I'm considering $k$ to be real. Then $a(k)$ and $b(k)$ against $k$ can be plotted on a graph. Let $c$ be the intersection value of $a(k)$ and $b(k)$ and $k_0$ the value of $k$ at which it occurs. It can be seen that the gradients of $a(k)$ and $b(k)$ will have the same sign. This restricts the possible situations where their signs are opposite. Namely if $|c|\leq1$ and $|k_0|\leq1$ then $a(k)$ and $b(k)$ will never have opposite signs for integer $k$ (feels true intuitively, I don't have rigorous argument for this). Solving for $k_0$ gives $k_0=\frac{d(a_0-b_0)}{x+y}$. From there $$\begin{align}|k_0|&\leq1\\\left|\frac{d(a_0-b_0)}{x+y}\right|&\leq 1\\d|(a_0-b_0)|&\leq|x+y|\end{align}$$

From there I got stuck as $a_0$ and $b_0$ are found through an algorithmic process so I can't think of how to manipulate them to prove that inequality. $|c|\leq1$ seems even harder to arrive at.

I'd appreciate if anyone can help prove or disprove my question. I'd also like to know if my method can be continued to find the answer.

Typo
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    Hint: when $\color{#c00}{xy<0}$ the line $,b = d/y \color{#c00}{- (x/y)},a,$ has positive $\rm\color{#c00}{slope}$ in $(a,b)$ plane. Can all of its integral points $(a,b)$ lie only in quadrant $1\ (a,b > 0)$ or quadrant $3\ (a,b < 0)?$ See here for some geometric intuition on such discrete lines. (it'd be better to use $x,y$ for the Bezout coef's so we'd have an $(x,y)$ plane). – Bill Dubuque Jun 20 '22 at 18:02
  • e.g. $,2a-3b = 1,$ has integral points $,(a,b) = ,\ldots(-4,-3),(-1,-1),(2,1),(5,3)\ldots$ (add or subtract $(3,2)$ to traverse them) – Bill Dubuque Jun 20 '22 at 18:27
  • @BillDubuque Seems I was needlessly complicating things by considering two lines instead of one. From the line you suggested, the $a$ and $b$ intercepts are $\frac dx$ and $\frac dy$ respectively. That pretty much bounds how much the line can extend into the 3rd quadrant, as in it can at most be $b=a-1$ in the case of $x=-y$. Not sure if this is the takeaway you intended because I don't understand some stuff in the link. Specifically what is meant by normalization and why multiples of $(-n,m)$ were added instead of $(\frac{-n}{d},\frac md)$ & why it was added to $(x,y)$ and not $(x_0,y_0)$. – Typo Jun 21 '22 at 00:27
  • Ahh, after multiple rereads I understand the argument (and answers to the 2 questions in my prev. comment) of the linked thread. That is incredibly clever and is certainly going to be a tool in my repertoire from now on. $\tag*{}$ Now that I've convinced myself of the answer (negative) to my original question, I'd like to hear the geometric argument you would have written for it kindly (assuming what I found is not the takeaway you intended). – Typo Jun 21 '22 at 01:21
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    If $f$ is ${\rm\color{#c00}{strictly\ increasing}}\ &\ f(0)!\ge! -1$ then $a!>!0\color{#c00} \Rightarrow f(a)!>!f(0)!\ge! \color{#0a0}{\bf -1},,$ so $f(a)!\ge! \color{#0a0}{\bf 0},$ if $f(a)\color{#0a0}{!\in! \Bbb Z}.,$ Thus $a,b$ have same sign for $a!>!0;,$ ditto if $a!<!0,$ & $f(0)\le 0$ by $f(a)!<!f(0)\le 0.,$ Apply that to $,b = f(a) = d/y \color{#c00}{- (x/y)}, a,,\ d!=!\gcd(x,y)!>! 0,,\ x!>!0!>!y\ $ (wlog) $\ \ $ – Bill Dubuque Jun 21 '22 at 11:43
  • Thanks very much! By the way, maybe you should post that as an answer so I can accept it? – Typo Jun 21 '22 at 11:57

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Hint: apply below to ($\rm\color{#0a0}{discrete}$) line $\,\overbrace{b = f(a) = d/y \color{#c00}{- (x/y)}\,a}^{\textstyle ax + by = d},\ d\!=\!\gcd(x,y)\!>\! 0,\,\ \color{#c00}{x\!>\!0\!>\!y}$

If $f$ is ${\rm\color{#c00}{strictly\ increasing}}\ \&\ f(0)\!\ge\! -1$ then $a\!>\!0\color{#c00} \Rightarrow f(a)\!>\!f(0)\!\ge\! \color{#0a0}{\bf -1},\,$ so $f(a)\!\ge\! \color{#0a0}{\bf 0}\,$ if $f(a)\color{#0a0}{\!\in\! \Bbb Z}.\,$ Thus $a,b$ have same sign if $a>0;\,$ ditto if $\,a\!<\!0\,$ & $f(0)\le 0,\,$ by $f(a)\!<\!f(0)\le 0.\,$

Bill Dubuque
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