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I have a question: Let $n\in\mathbb{Z}$, then $(a+bn,b)=(a,b)$, $(a,b)$ being the greatest common divisor.

Is this a correct approach?

Let $e=(a+bn,b), \quad n\in\mathbb{Z}$, then $e$ is expressible as the smallest linear combination of $an+b$ and $b$ so there exists $u,v\in\mathbb{Z}$ such that $$ e=u(a+bn)+vb $$ and $e$ is the smallest positive such number. Rearranging on the right hand side we get $$ e=ua+(un+v)b $$ which gives a linear combination of $a$ and $b$. Since $e$ is still the smallest such we get that $(a,b)=e$. If this is correct is there a more elegant solution?

Glorfindel
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Vinyl_cape_jawa
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    No, it's not correct because nothing tell you that $e$ is the smallest $e$ with the fact that $e=ua+(un+v)b$. Notice that you can suppose WLOG that $(a,b)=1$ and make your proof that will allow you to conclude. – Surb Jun 27 '17 at 11:26
  • @DietrichBurde: Proof-verification questions are rarely duplicates -- the answers to the older question may present a different proof but not tell what if anything there's wrong with this one. – hmakholm left over Monica Jun 27 '17 at 11:39
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    @HenningMakholm On the other hand the question also is "is there a more elegant solution?" I think this is the crucial part of the question, which therefore, I think, is basically a duplicate. Of course, one may have a different opinion here. – Dietrich Burde Jun 27 '17 at 11:46

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That doesn't sound quite convincing -- when you say "since $e$ is still the smallest such" it sounds to me like you're essentially assuming what you were supposed to prove.

I suspect your strategy is to show that the integer combinations of $a+bn$ and $b$ are the same as the integer combinations of $a$ and $b$. But in order to do that without handwaving you need to show that every combination of $a+bn$ and $b$ is also a combination of $a$ and $b$, and that every combination of $a$ and $b$ is also a combination of $a+bn$ and $b$. Your proof as it stands only does the first half of this.

(An alternative which seems more direct to me would be to show that the common divisors of $a+bn$ and $b$ are the same as the common divisors of $a$ and $b$).

hmakholm left over Monica
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  • Assume that $d=(a,b)$ and $e=(a+nb,b)$, then $$ d\mid a\qquad d\mid b $$ and therefore $$ d\mid bn $$ so $$ d\mid a+bn. $$ We see that $d\mid e$ since $d$ divides everything $e$ divides. Looking at the second assumption we can conclude that $$ e\mid b\qquad e\mid a+nb $$ therefore $$ e\mid a\qquad e\mid nb $$ this gives $d\mid e$. But $$ d\mid e, e\mid d\Rightarrow d=e. $$ – Vinyl_cape_jawa Jun 27 '17 at 11:44
  • something like this? – Vinyl_cape_jawa Jun 27 '17 at 11:44
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    @Vinyl_coat_jawa: That works (I'd want a bit more detail about how you conclude $e\mid a$, though), but was not exactly what I had in mind. I imagined a proof along the lines of "assume $x$ is some common divisor of $a$ and $a+nb$. Then (bla bla bla) so $x$ is also a common divisor of $a$ and $b$. And (bla bla bla) vice versa. Since being a common divisor of $a$ and $a+nb$ is the same property as being a common divisor of $a$ and $b$, the greatest number with this property must be the same no matter what we call the property." – hmakholm left over Monica Jun 27 '17 at 11:47
  • thank you!

    $e\mid a$ follows from $e\mid b$ and $e\mid a+nb$ right?

     – Vinyl_cape_jawa Jun 27 '17 at 11:49
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    @Vinyl_coat_jawa: Yes. – hmakholm left over Monica Jun 27 '17 at 12:07