Prove (or disprove) that $3^{2n+3}+40n-27$ is divisible by $64$ for any $n$ integer [duplicate]












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  • Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. [closed]

    4 answers




So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.






elementary-number-theory modular-arithmetic divisibility







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marked as duplicate by Bill Dubuque divisibility
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 Jan 26 at 1:04


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  • $begingroup$
    Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
    $endgroup$
    – lhf
    Jan 26 at 1:19






  • 1




    $begingroup$
    @lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
    $endgroup$
    – Bill Dubuque
    Jan 26 at 3:24


















0












$begingroup$



This question already has an answer here:




  • Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. [closed]

    4 answers




So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.






elementary-number-theory modular-arithmetic divisibility







share|cite|improve this question











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marked as duplicate by Bill Dubuque divisibility
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 Jan 26 at 1:04


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  • $begingroup$
    Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
    $endgroup$
    – lhf
    Jan 26 at 1:19






  • 1




    $begingroup$
    @lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
    $endgroup$
    – Bill Dubuque
    Jan 26 at 3:24
















0












0








0





$begingroup$



This question already has an answer here:




  • Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. [closed]

    4 answers




So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.






elementary-number-theory modular-arithmetic divisibility







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. [closed]

    4 answers




So I want to prove that is in title. I'm working with modular arithmetic, so if any of you can show me a hint, it will be perfect.





This question already has an answer here:




  • Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. [closed]

    4 answers







elementary-number-theory modular-arithmetic divisibility




elementary-number-theory modular-arithmetic divisibility






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share|cite|improve this question













share|cite|improve this question




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edited Jan 25 at 23:50









kccu

10.6k11229




10.6k11229










asked Jan 25 at 23:48









Richard KingstonRichard Kingston

283




283




marked as duplicate by Bill Dubuque divisibility
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 Jan 26 at 1:04


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marked as duplicate by Bill Dubuque divisibility
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 Jan 26 at 1:04


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
    $endgroup$
    – lhf
    Jan 26 at 1:19






  • 1




    $begingroup$
    @lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
    $endgroup$
    – Bill Dubuque
    Jan 26 at 3:24




















  • $begingroup$
    Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
    $endgroup$
    – lhf
    Jan 26 at 1:19






  • 1




    $begingroup$
    @lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
    $endgroup$
    – Bill Dubuque
    Jan 26 at 3:24


















$begingroup$
Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
$endgroup$
– lhf
Jan 26 at 1:19




$begingroup$
Hint: Let $x_n = 3^{2n+3}+40n-27$. Compute $x_{n+1}-x_{n} = 8 (3^{2 n + 3} + 5) bmod 64$.
$endgroup$
– lhf
Jan 26 at 1:19




1




1




$begingroup$
@lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
$endgroup$
– Bill Dubuque
Jan 26 at 3:24






$begingroup$
@lhf Easier to use $ x_{n+1}-color{#c00}9,x_n = 64(4-5n). $ This works generally - see this answer in the dupe.
$endgroup$
– Bill Dubuque
Jan 26 at 3:24












2 Answers
2






active

oldest

votes


















3












$begingroup$

We find that
$$
frac{3^{2n+3}+40n-27}{8}=frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+cdots+1)+5n.
$$ And also
$$
27(9^{n-1}+9^{n-2}+cdots+1)+5nequiv 27(1+1+cdots +1)+5nequiv32nequiv 0 ;text{(mod};8).
$$ Therefore $64 ;|; 3^{2n+3}+40n-27$ for all $nge 1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
    $endgroup$
    – Richard Kingston
    Jan 26 at 0:10










  • $begingroup$
    @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
    $endgroup$
    – Song
    Jan 26 at 0:12





















0












$begingroup$

Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27equiv 0mod64$. We have to show that
$$3^{2n+5}+40(n+1)-27=9cdot3^{2n+3}+40n+13equiv 0mod64.$$
The inductive hypothesis can be rewritten as
$$3^{2n+3}equiv -40n +27mod64,$$
so that
$$9cdot3^{2n+3}+40n+13equiv-360n+243+40m+13=-320n+256equiv 0mod 64. $$






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    We find that
    $$
    frac{3^{2n+3}+40n-27}{8}=frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+cdots+1)+5n.
    $$     And also
    $$
    27(9^{n-1}+9^{n-2}+cdots+1)+5nequiv 27(1+1+cdots +1)+5nequiv32nequiv 0 ;text{(mod};8).
    $$     Therefore $64 ;|; 3^{2n+3}+40n-27$ for all $nge 1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
      $endgroup$
      – Richard Kingston
      Jan 26 at 0:10










    • $begingroup$
      @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
      $endgroup$
      – Song
      Jan 26 at 0:12


















    3












    $begingroup$

    We find that
    $$
    frac{3^{2n+3}+40n-27}{8}=frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+cdots+1)+5n.
    $$     And also
    $$
    27(9^{n-1}+9^{n-2}+cdots+1)+5nequiv 27(1+1+cdots +1)+5nequiv32nequiv 0 ;text{(mod};8).
    $$     Therefore $64 ;|; 3^{2n+3}+40n-27$ for all $nge 1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
      $endgroup$
      – Richard Kingston
      Jan 26 at 0:10










    • $begingroup$
      @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
      $endgroup$
      – Song
      Jan 26 at 0:12
















    3












    3








    3





    $begingroup$

    We find that
    $$
    frac{3^{2n+3}+40n-27}{8}=frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+cdots+1)+5n.
    $$     And also
    $$
    27(9^{n-1}+9^{n-2}+cdots+1)+5nequiv 27(1+1+cdots +1)+5nequiv32nequiv 0 ;text{(mod};8).
    $$     Therefore $64 ;|; 3^{2n+3}+40n-27$ for all $nge 1$.






    share|cite|improve this answer









    $endgroup$



    We find that
    $$
    frac{3^{2n+3}+40n-27}{8}=frac{27(9^n-1)+40n}{8}=27(9^{n-1}+9^{n-2}+cdots+1)+5n.
    $$     And also
    $$
    27(9^{n-1}+9^{n-2}+cdots+1)+5nequiv 27(1+1+cdots +1)+5nequiv32nequiv 0 ;text{(mod};8).
    $$     Therefore $64 ;|; 3^{2n+3}+40n-27$ for all $nge 1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 26 at 0:08









    SongSong

    17.4k21246




    17.4k21246












    • $begingroup$
      i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
      $endgroup$
      – Richard Kingston
      Jan 26 at 0:10










    • $begingroup$
      @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
      $endgroup$
      – Song
      Jan 26 at 0:12




















    • $begingroup$
      i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
      $endgroup$
      – Richard Kingston
      Jan 26 at 0:10










    • $begingroup$
      @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
      $endgroup$
      – Song
      Jan 26 at 0:12


















    $begingroup$
    i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
    $endgroup$
    – Richard Kingston
    Jan 26 at 0:10




    $begingroup$
    i understand, so your answer is reducible to say "if is divisible by 8, too is divisible by 8^2" ?
    $endgroup$
    – Richard Kingston
    Jan 26 at 0:10












    $begingroup$
    @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
    $endgroup$
    – Song
    Jan 26 at 0:12






    $begingroup$
    @RichardKingston Basically, yes. If $8 | N$ and $8 | (N/8)$, then it holds $64 | N$ .
    $endgroup$
    – Song
    Jan 26 at 0:12













    0












    $begingroup$

    Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27equiv 0mod64$. We have to show that
    $$3^{2n+5}+40(n+1)-27=9cdot3^{2n+3}+40n+13equiv 0mod64.$$
    The inductive hypothesis can be rewritten as
    $$3^{2n+3}equiv -40n +27mod64,$$
    so that
    $$9cdot3^{2n+3}+40n+13equiv-360n+243+40m+13=-320n+256equiv 0mod 64. $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27equiv 0mod64$. We have to show that
      $$3^{2n+5}+40(n+1)-27=9cdot3^{2n+3}+40n+13equiv 0mod64.$$
      The inductive hypothesis can be rewritten as
      $$3^{2n+3}equiv -40n +27mod64,$$
      so that
      $$9cdot3^{2n+3}+40n+13equiv-360n+243+40m+13=-320n+256equiv 0mod 64. $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27equiv 0mod64$. We have to show that
        $$3^{2n+5}+40(n+1)-27=9cdot3^{2n+3}+40n+13equiv 0mod64.$$
        The inductive hypothesis can be rewritten as
        $$3^{2n+3}equiv -40n +27mod64,$$
        so that
        $$9cdot3^{2n+3}+40n+13equiv-360n+243+40m+13=-320n+256equiv 0mod 64. $$






        share|cite|improve this answer









        $endgroup$



        Use induction and modular arithmetic: suppose that, for some $n$, $3^{2n+3}+40n-27equiv 0mod64$. We have to show that
        $$3^{2n+5}+40(n+1)-27=9cdot3^{2n+3}+40n+13equiv 0mod64.$$
        The inductive hypothesis can be rewritten as
        $$3^{2n+3}equiv -40n +27mod64,$$
        so that
        $$9cdot3^{2n+3}+40n+13equiv-360n+243+40m+13=-320n+256equiv 0mod 64. $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 0:22









        BernardBernard

        122k741116




        122k741116















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