It is well known that the equation $ax+by = n$ has a solution in integers $x$ and $y$ iff $gcd(a,b)|n$. I would like to know if there is a theorem in number theory to show that there is a relationship between $x$ and $y$. An example of such a relationship might be their being relatively prime or something similar to that.
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$x$ and $y$ are not unique. There are infinitely many solutions $x,y$ if $d\mid n$, see here. – Dietrich Burde Nov 04 '16 at 19:18
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I know that , but i seek for example if there is known theorem show the existence of relationship between those solutions – zeraoulia rafik Nov 04 '16 at 19:18
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Converse of Bezout's Theorem, perhaps. – SchrodingersCat Nov 04 '16 at 19:19
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There's the obvious relationship that $y=(ax-n)/b$... – Matt B Nov 04 '16 at 19:20
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but this can't tel us any thing about the nature of the existed relationship between x, and y in the view of elementary number theory , check out the answer of :Dietrich Burde – zeraoulia rafik Nov 04 '16 at 19:33
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you might be interested about Popovicius' Theorem, have a look at this paper – G Cab Nov 04 '16 at 20:07
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Given $a, b, n$ fixed, there always is an explicit description of all the pairs $(x, y)$. See e.g. here. – Pavel Čoupek Nov 04 '16 at 20:07
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Thanks for the slick link , i have to check it – zeraoulia rafik Nov 04 '16 at 20:13
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@PavelČoupek, i don't know no vote neither question and neither answer and comments i don't know if it is a trivial question or no interest of people for it !!! – zeraoulia rafik Nov 04 '16 at 21:03
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the matter is that the solution to linear diophantine equations is an argument that has been dealt since .. Diophantus, and there is a vast literature about. So, unless you have a more specific question, there is no much to say, that you cannot find on the web. – G Cab Nov 04 '16 at 21:49
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I think i specified "An example of such a relationship might be their being relatively prime or something similar to that." – zeraoulia rafik Nov 04 '16 at 21:51
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If $ax+by=d\mid n$ for $d=gcd(a,b)$, then we must have $gcd(x,y)=1$. So in this case the solutions $x$ and $y$ are relatively prime, see this MSE-question.
Dietrich Burde
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thanks a lot this what i feel and what i wrote in my question !!! – zeraoulia rafik Nov 04 '16 at 19:25
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I have checked ur link but in the linked question instead n he cited gcd(a,b) and the last is not necessary devides n – zeraoulia rafik Nov 04 '16 at 19:42
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But $d=gcd(a,b)$ necessarily divides $n$. You have written it yourself in the question, because otherwise there is no solution $x,y$. – Dietrich Burde Nov 04 '16 at 19:52
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mayeb i can understanding it as this : if it has a solution then gcd(x,y)=1 – zeraoulia rafik Nov 04 '16 at 19:53
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Vote it's not important for me only to improve my mathematical skill's – zeraoulia rafik Nov 04 '16 at 19:57
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Consider any pair $(x_0,y_0)$ and any pair $(a,b).$ Define $n=ax_0+by_0.$ We have that $(x_0,y_0)$ is a solution of $ax+by=n.$ Since the pair $(x_0,y_0)$ is arbitrary you can't expect any relation in general.
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What we can say is that $gcd(x_0,y_0)|n.$ Why? Because the equation $x_0x+y_0y=n$ has solution: $(a,b).$
mfl
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Not always there are solutions because if $(a,b)=d\gt 1$ then it is clear that $d$ must divide $n$. When there is a solution $(x_0,y_0)$ you have $$\begin{cases}ax+by=n\\ax_0+by_0=n\end{cases}\Rightarrow a(x-x_0)+b(y-y_0)=0\Rightarrow\begin{cases}x-x_0=bt\\y-y_0=at\end{cases}$$ which gives a parametrization of $x$ and $y$ in function of $t$.
Piquito
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