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











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










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  • Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
    – lulu
    Nov 19 at 18:11










  • 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.
    – Acccumulation
    Nov 19 at 19:00










  • @Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
    – Ovi
    Nov 19 at 20:03

















up vote
1
down vote

favorite
1












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










share|cite|improve this question
























  • Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
    – lulu
    Nov 19 at 18:11










  • 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.
    – Acccumulation
    Nov 19 at 19:00










  • @Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
    – Ovi
    Nov 19 at 20:03















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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










share|cite|improve this question















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







abstract-algebra algebra-precalculus






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edited Nov 19 at 18:15









greedoid

35.6k114590




35.6k114590










asked Nov 19 at 18:10









ten1o

1335




1335












  • Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
    – lulu
    Nov 19 at 18:11










  • 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.
    – Acccumulation
    Nov 19 at 19:00










  • @Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
    – Ovi
    Nov 19 at 20:03




















  • Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
    – lulu
    Nov 19 at 18:11










  • 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.
    – Acccumulation
    Nov 19 at 19:00










  • @Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
    – Ovi
    Nov 19 at 20:03


















Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
Nov 19 at 18:11




Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
Nov 19 at 18:11












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.
– Acccumulation
Nov 19 at 19:00




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.
– Acccumulation
Nov 19 at 19:00












@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
Nov 19 at 20:03






@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
Nov 19 at 20:03












2 Answers
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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 $n=7k+4$.






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  • 2




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


















up vote
1
down vote













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






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    2 Answers
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    up vote
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    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 $n=7k+4$.






    share|cite|improve this answer



















    • 2




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















    up vote
    1
    down vote













    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 $n=7k+4$.






    share|cite|improve this answer



















    • 2




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













    up vote
    1
    down vote










    up vote
    1
    down vote









    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 $n=7k+4$.






    share|cite|improve this answer














    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 $n=7k+4$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 19 at 19:59

























    answered Nov 19 at 18:14









    greedoid

    35.6k114590




    35.6k114590








    • 2




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














    • 2




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








    2




    2




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




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










    up vote
    1
    down vote













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






    share|cite|improve this answer



























      up vote
      1
      down vote













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






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









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






        share|cite|improve this answer














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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 20:08

























        answered Nov 19 at 18:38









        Bill Dubuque

        207k29189624




        207k29189624






























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