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  • Q1: Prove that $gcd(k,n+k)=1$ if and only if $gcd(n,k)=1$

If the $\gcd(k,n+k)=1$ then by Bézout's identity there exists integers $x$ and $y$ such that $kx+(n+k)y=1$.

$kx+(n+k)y=kx+ny+ky=k(x+y)+ny=1$ therefore $\gcd(n,k)=1$ by Bézout's identity (Given integers $a$,$b$, not both $0$, there exist integers $x,y$ such that $ax+by=\gcd(a,b)$)(we know $x+y$ is an integer because the sum of two integers is also an integer)

For the converse, if the $\gcd(n,k)=1$ then there exists integers $x$ and $y$ such that $nx+ky=1$. $nx+ky=nx+kx-kx+ky=(n+k)x+k(y-x)=1$, therefore $\gcd(k,n+k)=1$ by Bézout's identity agian.

  • Q2: Is it true that $\text{gcd}(k,n+k)=d$ if and only if $\text{gcd}(k,n)=d$

I said true and just replaced the $1$ in the previous proof with $d$, thought I'm not sure this is true.

  • Q3: Let $a,b,c,d$ be strictly positive integers. Prove that if $a$ divides $b$ and $b$ divides $c$ and $\gcd(a,c)=1$ then we must have $a=1$.

I just used basic divisibility properties (one's which we've already proved) for this one, $a$ divides $b$ and $b$ divides $c$ therefore $a$ divides $c$, therefore $a=\gcd(a,c)$ and hence $a=1$.

Uncertain about a lot of stuff here (whether what I've put is true/actually proves the statements), thanks in advance for any help!

Yajat Shamji
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Questioner
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    please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Oct 22 '18 at 16:55
  • You don't need Bézout's identity: if $x\mid k$ and $x\mid(n+k)$, then $x\mid k$ and $x\mid n$; conversely, if $x\mid k$ and $x\mid n$, then $x\mid k$ and $x\mid (n+k)$. This answers both 1 and 2. – egreg Oct 22 '18 at 17:22

1 Answers1

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$(1)$ is fine. For $(2)$ you need to do more, e.g. show that both gcds divide each other. More simply your equations imply the set equality $\ k\Bbb Z + (n\!+\!k)\Bbb Z\, =\, k\Bbb Z + n\Bbb Z,\,$ implying equality of their least positive elements, which means $\,\gcd(k,\,n\!+\!k)\, =\, \gcd(k,n).\,$ Finally $(3)$ is fine.

Note $ $ For $(1)$ you might need to say more depending on the particular statement of Bezout's Identity in your textbook, i.e. you might need to finish with $\,d\mid n,k\Rightarrow\, d\mid k(x\!+\!y)+ny = 1$

It is more natural (and general) to prove $(1)$ and $(2)$ by divisibility laws rather than Bezout, i.e.

note that If $\ d\mid k\ $ then $\ d\mid mk \!+\! n \iff d\mid n.\, $ Therefore $\,k,\, mk\!+\!n\,$ and $\,k,\, n$ have exactly the same set $\,S\,$ of common divisors $\,d,\,$ hence they have the same greatest common divisor $(= \max S)$

Said $\!\bmod d\!:\, $ if $\ k\equiv 0\ $ then $\ mk\!+\!n\equiv 0 \iff n\equiv 0$

Bill Dubuque
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