I have got 2 questions which I could not solve:

1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$

2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$

  • $\begingroup$ Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas. $\endgroup$ – lulu Nov 19 '18 at 18:11
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    $\begingroup$ When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer. $\endgroup$ – Acccumulation Nov 19 '18 at 19:00
  • $\begingroup$ @Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$" $\endgroup$ – Ovi Nov 19 '18 at 20:03

Let $d=\gcd(8n+6,6n+3)$, then $$d\mid 8n+6$$

$$d\mid 6n+3$$

so $$d\mid 6(8n+6)-8(6n+3)= 12$$

so $d\in \{1,2,3,4,6,12\}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)

For second one:

Let $d=\gcd(2n^2+3n+5,n^2 + n+1)$, then $$d\mid 2n^2+3n+5$$

$$d\mid n^2+n+1$$

so $$d\mid 2n^2+3n+5-2(n^2 + n+1) =n+3$$

then $$d\mid (n^2+n+1)-(n^2-9)-(n+3)=7$$

So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.

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    $\begingroup$ Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same $\endgroup$ – Mark Bennet Nov 19 '18 at 18:34

$(1)$ A euclidean sequence is $\ \overbrace{8n\!+\!6,\,6n\!+\!3,\,2n\!+\!3,\,{-}\color{#c00}6}^{\Large a_{k-1} -\, j\ a_k\ =\ a_{k+1}}\,$ so the gcd is

$$(2n\!+\!3,\,\color{#c00}{2\cdot 3}) = (2n\!+\!3,\color{#c00}2)(2n\!+\!3,\color{#c00}3) = (3,2)(2n,3) = (n,3)\qquad\qquad $$

$(2)$ A euclidean sequence is $\ 2n^2\!+\!3n\!+\!5,\!\!\!\!\underbrace{n^2\!+\!n\!+\!1,\, n\!+\!3,\, \color{#0a0}7}_{\large f(n)\ \equiv\ \color{#0a0}{f(-3)}\,\pmod{\!n+3}}\!\!\!\!$ so the gcd $= (n\!+\!3,7)$


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