1
$\begingroup$

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 $(2n^2+3n+5,n^2+n+1)$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $n=7k+4$.

$\endgroup$
  • 2
    $\begingroup$ 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
$\begingroup$

$(1)$ 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 $= (n\!+\!3,7)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.