Ytukyg

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

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

$$dmid 6n+3$$

so $$dmid 6(8n+6)-8(6n+3)= 12$$

so $din {1,2,3,4,6,12}$ 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 $$dmid 2n^2+3n+5$$

$$dmid n^2+n+1$$

so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$

then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$

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

$(1)$ 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

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

$(2)$ 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 $= (n!+!3,7)$