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I was asked to determine all possible couples $(a,b) \in \mathbb{Z}^2$ which satisfies $\text{gcd}(411,2016) = a411 + b2016$. Using the Euclidean division alogirthm, I have found that \begin{align} 2016 &= 4 \cdot 411 + 372\\ 411 &= 1 \cdot 372 + 39\\ 372 &= 9 \cdot 39 + 21\\ 39 &= 1 \cdot 21 + 18\\ 21 &= 1 \cdot 18 + 3\\ 18 &= 3 \cdot 6 \end{align} from which it folows that $\text{gcd}(411,2016) = 3$. Working back, I find that $3 = 21 \cdot 2016 - 103 \cdot 411$.

In order to determine the other solutions, I worked more generale: suppose we want to find $x,y \in \mathbb{Z}$ such that $ax + by = d$. Suppose that $(x,y), (x',y')$ are two solutions, then filling this in we find that $a(x-x') = b(y' - y)$. Supposing that $a,b$ are coprime (as is the case for my expansion of $3$, we must have that $a$ is a divisor of $y'-y$ and $b$ is a divisor of $x-x'$. Therefore, there are $k, k' \in \mathbb{Z}$ such that $ak = y'-y$ and $bk' = x-x'$. However, I have no idea how $k,k'$ are related. Any hints?

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    Note there is an *Extended Euclidean algorithm* which avoids having to go backwards, and in addition yields the values of $\dfrac a{a\wedge b}$ and $\dfrac b{a\wedge b}$. – 2017-02-14
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    I Will have to look this up. Thanks for your hint, since this procedure felt quite awkward, so I already wondered if there was not a shorter way to do this. – 2017-02-14
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    You may take a look at my answer to [this question](http://math.stackexchange.com/questions/2099165/how-to-solve-min-x-in-mathbb-n-0-quad-x-cdot-714-quad-mod-quad-1972-qua/2099198#2099198) if you want to see the very simple layout of the alogorithm (to have the quotients by the g.c.d., one further step is required). – 2017-02-14

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Suppose we have $\gcd(a,b)=d$, and write $a'=a/d$, $b'=b/d$. Now $a'$ and $b'$ are coprime. If $ax+by=ax'+by'=d$, we must have $a'x+b'y=a'x'+b'y'=1$. Now $a'(x-x')=b'(y'-y)$, so $b'|a'(x-x')$, and since $b',a'$ are coprime we must have $b'|(x-x')$. Writing $x'=x+nb'$, we get $y'=y-na'$. So the general solution for pairs with $ax'+by'=d$ is $(x+nb/d,y-na/d)$ for $n\in\mathbb Z$.

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    it is so obvious... I have no idea why I did not see this! – 2017-02-14
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    @Student It is even more obvious using well-known properties of fractions - see my answer. (Note: I updated the link in my answer to a better one, which includes showing how this can be used very slickly in irrationality proofs of square-roots - one of John Conway's favorite ways). – 2017-02-14
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Hint $ $ Suppose $\, a/b = a'/b'\,$ where $a'/b'$ is reduced (in lowest terms). Then we have

$$ \begin{array}{}\large \frac{y'-y}{x'-x}\, =\, \frac{a}b\, =\, \frac{\color{#0a0}{a'}}{\color{#c00}{b'}}\\[.3em] \Rightarrow\,\ \begin{align}{y'-y} \:&=\: n\,\color{#0a0}{a'}\\ x'-x \:&=\: n\,\color{#c00}{b'}\end{align} \end{array}$$

for some integer $n,\,$ by a basic principality property of reduced fractions (Unique Fractionization)

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    Thank you very much for your answer and your comment. These have been very helpful! – 2017-02-14
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This is a linear Diophantine equation. Have a look here or here.

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    Thank you for your answer! I have never heard about Diophantine equatins, but I think that it is worth getting to know them. Really appreciate your help! – 2017-02-14
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    My thought was that you compare the solution algorithm from the links with your effort. – 2017-02-15