$5nmid n(n+1)implies 5mid (n^3-6n^2+n-1)$












0












$begingroup$


Prove $5nmid n(n+1)implies 5mid (n^3-6n^2+n-1)$. My try:



$5nmid n(n+1)implies 5mid (n-2)(n-1)(n+2)=n^3-n^2-4n+4implies$
$5mid (n^3-n^2-4n+4)-(5n^2+ 5(n-1))=n^3-6n^2+n-1$.



This time a simple trick worked, but how to solve this (kind of) problem more methodically, maybe using modular arithmetic?






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
    $endgroup$
    – Doug M
    Jan 11 at 21:34












  • $begingroup$
    Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
    $endgroup$
    – fleablood
    Jan 11 at 22:38










  • $begingroup$
    Yes @fleablood ut I believe modular arithmetic is more than that.
    $endgroup$
    – Lehs
    Jan 11 at 22:44
















0












$begingroup$


Prove $5nmid n(n+1)implies 5mid (n^3-6n^2+n-1)$. My try:



$5nmid n(n+1)implies 5mid (n-2)(n-1)(n+2)=n^3-n^2-4n+4implies$
$5mid (n^3-n^2-4n+4)-(5n^2+ 5(n-1))=n^3-6n^2+n-1$.



This time a simple trick worked, but how to solve this (kind of) problem more methodically, maybe using modular arithmetic?






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
    $endgroup$
    – Doug M
    Jan 11 at 21:34












  • $begingroup$
    Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
    $endgroup$
    – fleablood
    Jan 11 at 22:38










  • $begingroup$
    Yes @fleablood ut I believe modular arithmetic is more than that.
    $endgroup$
    – Lehs
    Jan 11 at 22:44














0












0








0





$begingroup$


Prove $5nmid n(n+1)implies 5mid (n^3-6n^2+n-1)$. My try:



$5nmid n(n+1)implies 5mid (n-2)(n-1)(n+2)=n^3-n^2-4n+4implies$
$5mid (n^3-n^2-4n+4)-(5n^2+ 5(n-1))=n^3-6n^2+n-1$.



This time a simple trick worked, but how to solve this (kind of) problem more methodically, maybe using modular arithmetic?






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




Prove $5nmid n(n+1)implies 5mid (n^3-6n^2+n-1)$. My try:



$5nmid n(n+1)implies 5mid (n-2)(n-1)(n+2)=n^3-n^2-4n+4implies$
$5mid (n^3-n^2-4n+4)-(5n^2+ 5(n-1))=n^3-6n^2+n-1$.



This time a simple trick worked, but how to solve this (kind of) problem more methodically, maybe using modular arithmetic?






algebra-precalculus elementary-number-theory modular-arithmetic




algebra-precalculus elementary-number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 21:19









LehsLehs

7,02831662




7,02831662








  • 2




    $begingroup$
    I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
    $endgroup$
    – Doug M
    Jan 11 at 21:34












  • $begingroup$
    Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
    $endgroup$
    – fleablood
    Jan 11 at 22:38










  • $begingroup$
    Yes @fleablood ut I believe modular arithmetic is more than that.
    $endgroup$
    – Lehs
    Jan 11 at 22:44














  • 2




    $begingroup$
    I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
    $endgroup$
    – Doug M
    Jan 11 at 21:34












  • $begingroup$
    Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
    $endgroup$
    – fleablood
    Jan 11 at 22:38










  • $begingroup$
    Yes @fleablood ut I believe modular arithmetic is more than that.
    $endgroup$
    – Lehs
    Jan 11 at 22:44








2




2




$begingroup$
I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
$endgroup$
– Doug M
Jan 11 at 21:34






$begingroup$
I like your approach. If you want to be methodical, brute force the cases. $n = 1, n=2, n=3$ and for any polynomial if $a|f(n)$ then $a|f(n+a)$ which can be shown using modular arithmetic or by the properties of binomials.
$endgroup$
– Doug M
Jan 11 at 21:34














$begingroup$
Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
$endgroup$
– fleablood
Jan 11 at 22:38




$begingroup$
Well being able to say $5not mid (n-2)(1-1)(n+2)$ is modular arithmetic. Otherwise how can you justify it?
$endgroup$
– fleablood
Jan 11 at 22:38












$begingroup$
Yes @fleablood ut I believe modular arithmetic is more than that.
$endgroup$
– Lehs
Jan 11 at 22:44




$begingroup$
Yes @fleablood ut I believe modular arithmetic is more than that.
$endgroup$
– Lehs
Jan 11 at 22:44










4 Answers
4






active

oldest

votes


















3












$begingroup$

$$ n^3 - 6 n^2 + n - 1 equiv n^3 - n^2 + n - 1 equiv (n^2 + 1)(n-1) pmod 5 $$



The hypothesis says that $n neq 0,4 pmod 5,$ so that $n equiv 1,2,3 pmod 5.$ Both $2^2 + 1 equiv 3^2 + 1 equiv 0 pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 pmod 5$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
    $endgroup$
    – N. S.
    Jan 11 at 23:49










  • $begingroup$
    @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
    $endgroup$
    – Will Jagy
    Jan 11 at 23:59










  • $begingroup$
    @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
    $endgroup$
    – Bill Dubuque
    Jan 12 at 20:40



















1












$begingroup$

Not really answering the question, I believe, but ... here is another way
$$5 mid n(n+1)(n+2)(n+3)(n+4)$$
since $5$ is prime and using Euclid's lemma
$$5 nmid n(n+1) Rightarrow \
5 mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 Rightarrow \
5 mid -1+n-6n^2+n^2$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $!bmod 5!:, f(n)equiv n^2(overbrace{n!-!1}^{Large n - 6})+n!-!1, =, (color{#c00}{n!-!1}) (overbrace{n^2+1}^{Largecolor{#0a0}{n^2 - 4}})$



    therefore we conclude $ underbrace{ nnotequiv 0,-1}_{Large 5 nmid n(n+1)}Rightarrow color{#c00}{nequiv 1},$ or $!!!!underbrace{nequiv pm2}_{Large Rightarrow color{#0a0}{n^2 equiv 4} }!!!!$ so $,f(n)equiv 0$




    Or: $, (color{#c00}{n+1}),f(n)equiv {n^4-1equiv 0 {rm by} nnotequiv 0},$ and little Fermat



    so $,color{#c00}{nnotequiv -1}Rightarrow f(n)equiv 0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
      $endgroup$
      – Bill Dubuque
      Jan 11 at 22:27





















    0












    $begingroup$

    $n^3 - 6n^2 + n-1 equiv n^3 - n^2 + n-1 pmod 5equiv$



    $(n^2 + 1)(n-1) equiv (n^2 -4)(n-1) pmod 5$



    $(n-2)(n+2)(n-1) pmod(5)$.



    If $n equiv 2,-2, 1 pmod 5$ then $n^3 - 5n^2 + n-1 equiv 0 pmod 5$ and $5|n^3 - 5n^2 + n-1$.



    As $5not mid n(n+1)$ and $5$ is prie $5not mid n$ and $5not mid n+1$ and $n not equiv 0pmod 5$ and $nnot equiv -1 pmod 5$.



    ${-2, -1, 0, 1, 2}$ is a complete residue class of $5$ so $nequiv -2,-1, 0, 1, $ or $2 pmod 5$ and since $nnot equiv 0,-1$ then $n equiv -2, 2$ or $1 pmod 5$. So we are done.



    ....



    Alternatively (now that we know where we want to go)



    $nnot mid n(n+1) implies$



    $n not mid 0$ or $-1pmod 5implies$



    $n mid 1,2,$ or $-2pmod 5implies$



    One of $n-1, n-2, n+2equiv 0 pmod 5implies$



    $(n-1)(n-2)(n+2)equiv 0 pmod 5 implies$



    $ n^3 -n^2 - 4n + 4equiv 0 pmod 5implies$



    $n^3 - 6n^2 + n - 1equiv 0 pmod 5 implies $



    $5|n^3 - 6n^2 + n-1$



    ....



    Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.



    $n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      $$ n^3 - 6 n^2 + n - 1 equiv n^3 - n^2 + n - 1 equiv (n^2 + 1)(n-1) pmod 5 $$



      The hypothesis says that $n neq 0,4 pmod 5,$ so that $n equiv 1,2,3 pmod 5.$ Both $2^2 + 1 equiv 3^2 + 1 equiv 0 pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 pmod 5$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
        $endgroup$
        – N. S.
        Jan 11 at 23:49










      • $begingroup$
        @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
        $endgroup$
        – Will Jagy
        Jan 11 at 23:59










      • $begingroup$
        @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
        $endgroup$
        – Bill Dubuque
        Jan 12 at 20:40
















      3












      $begingroup$

      $$ n^3 - 6 n^2 + n - 1 equiv n^3 - n^2 + n - 1 equiv (n^2 + 1)(n-1) pmod 5 $$



      The hypothesis says that $n neq 0,4 pmod 5,$ so that $n equiv 1,2,3 pmod 5.$ Both $2^2 + 1 equiv 3^2 + 1 equiv 0 pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 pmod 5$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
        $endgroup$
        – N. S.
        Jan 11 at 23:49










      • $begingroup$
        @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
        $endgroup$
        – Will Jagy
        Jan 11 at 23:59










      • $begingroup$
        @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
        $endgroup$
        – Bill Dubuque
        Jan 12 at 20:40














      3












      3








      3





      $begingroup$

      $$ n^3 - 6 n^2 + n - 1 equiv n^3 - n^2 + n - 1 equiv (n^2 + 1)(n-1) pmod 5 $$



      The hypothesis says that $n neq 0,4 pmod 5,$ so that $n equiv 1,2,3 pmod 5.$ Both $2^2 + 1 equiv 3^2 + 1 equiv 0 pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 pmod 5$






      share|cite|improve this answer









      $endgroup$



      $$ n^3 - 6 n^2 + n - 1 equiv n^3 - n^2 + n - 1 equiv (n^2 + 1)(n-1) pmod 5 $$



      The hypothesis says that $n neq 0,4 pmod 5,$ so that $n equiv 1,2,3 pmod 5.$ Both $2^2 + 1 equiv 3^2 + 1 equiv 0 pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 pmod 5$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 11 at 21:52









      Will JagyWill Jagy

      103k5101200




      103k5101200












      • $begingroup$
        $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
        $endgroup$
        – N. S.
        Jan 11 at 23:49










      • $begingroup$
        @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
        $endgroup$
        – Will Jagy
        Jan 11 at 23:59










      • $begingroup$
        @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
        $endgroup$
        – Bill Dubuque
        Jan 12 at 20:40


















      • $begingroup$
        $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
        $endgroup$
        – N. S.
        Jan 11 at 23:49










      • $begingroup$
        @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
        $endgroup$
        – Will Jagy
        Jan 11 at 23:59










      • $begingroup$
        @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
        $endgroup$
        – Bill Dubuque
        Jan 12 at 20:40
















      $begingroup$
      $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
      $endgroup$
      – N. S.
      Jan 11 at 23:49




      $begingroup$
      $$n^2-1 equiv (n-2)(n-3) pmod{5}$$ ;)
      $endgroup$
      – N. S.
      Jan 11 at 23:49












      $begingroup$
      @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
      $endgroup$
      – Will Jagy
      Jan 11 at 23:59




      $begingroup$
      @N.S. sure. That is the sort of thing where I have the conclusion memorized, square roots of -1 and so forth, on account of working with integer coefficient quadratic forms.
      $endgroup$
      – Will Jagy
      Jan 11 at 23:59












      $begingroup$
      @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
      $endgroup$
      – Bill Dubuque
      Jan 12 at 20:40




      $begingroup$
      @N.S. Simpler: $,5mid (pm2)^2!+1 $ Using a balanced (least magnitude) residue system often simplifies (manual) computations (as I attempted to highlight in my answer).
      $endgroup$
      – Bill Dubuque
      Jan 12 at 20:40











      1












      $begingroup$

      Not really answering the question, I believe, but ... here is another way
      $$5 mid n(n+1)(n+2)(n+3)(n+4)$$
      since $5$ is prime and using Euclid's lemma
      $$5 nmid n(n+1) Rightarrow \
      5 mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 Rightarrow \
      5 mid -1+n-6n^2+n^2$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Not really answering the question, I believe, but ... here is another way
        $$5 mid n(n+1)(n+2)(n+3)(n+4)$$
        since $5$ is prime and using Euclid's lemma
        $$5 nmid n(n+1) Rightarrow \
        5 mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 Rightarrow \
        5 mid -1+n-6n^2+n^2$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Not really answering the question, I believe, but ... here is another way
          $$5 mid n(n+1)(n+2)(n+3)(n+4)$$
          since $5$ is prime and using Euclid's lemma
          $$5 nmid n(n+1) Rightarrow \
          5 mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 Rightarrow \
          5 mid -1+n-6n^2+n^2$$






          share|cite|improve this answer









          $endgroup$



          Not really answering the question, I believe, but ... here is another way
          $$5 mid n(n+1)(n+2)(n+3)(n+4)$$
          since $5$ is prime and using Euclid's lemma
          $$5 nmid n(n+1) Rightarrow \
          5 mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 Rightarrow \
          5 mid -1+n-6n^2+n^2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 22:00









          rtybasertybase

          10.7k21533




          10.7k21533























              1












              $begingroup$

              $!bmod 5!:, f(n)equiv n^2(overbrace{n!-!1}^{Large n - 6})+n!-!1, =, (color{#c00}{n!-!1}) (overbrace{n^2+1}^{Largecolor{#0a0}{n^2 - 4}})$



              therefore we conclude $ underbrace{ nnotequiv 0,-1}_{Large 5 nmid n(n+1)}Rightarrow color{#c00}{nequiv 1},$ or $!!!!underbrace{nequiv pm2}_{Large Rightarrow color{#0a0}{n^2 equiv 4} }!!!!$ so $,f(n)equiv 0$




              Or: $, (color{#c00}{n+1}),f(n)equiv {n^4-1equiv 0 {rm by} nnotequiv 0},$ and little Fermat



              so $,color{#c00}{nnotequiv -1}Rightarrow f(n)equiv 0$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
                $endgroup$
                – Bill Dubuque
                Jan 11 at 22:27


















              1












              $begingroup$

              $!bmod 5!:, f(n)equiv n^2(overbrace{n!-!1}^{Large n - 6})+n!-!1, =, (color{#c00}{n!-!1}) (overbrace{n^2+1}^{Largecolor{#0a0}{n^2 - 4}})$



              therefore we conclude $ underbrace{ nnotequiv 0,-1}_{Large 5 nmid n(n+1)}Rightarrow color{#c00}{nequiv 1},$ or $!!!!underbrace{nequiv pm2}_{Large Rightarrow color{#0a0}{n^2 equiv 4} }!!!!$ so $,f(n)equiv 0$




              Or: $, (color{#c00}{n+1}),f(n)equiv {n^4-1equiv 0 {rm by} nnotequiv 0},$ and little Fermat



              so $,color{#c00}{nnotequiv -1}Rightarrow f(n)equiv 0$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
                $endgroup$
                – Bill Dubuque
                Jan 11 at 22:27
















              1












              1








              1





              $begingroup$

              $!bmod 5!:, f(n)equiv n^2(overbrace{n!-!1}^{Large n - 6})+n!-!1, =, (color{#c00}{n!-!1}) (overbrace{n^2+1}^{Largecolor{#0a0}{n^2 - 4}})$



              therefore we conclude $ underbrace{ nnotequiv 0,-1}_{Large 5 nmid n(n+1)}Rightarrow color{#c00}{nequiv 1},$ or $!!!!underbrace{nequiv pm2}_{Large Rightarrow color{#0a0}{n^2 equiv 4} }!!!!$ so $,f(n)equiv 0$




              Or: $, (color{#c00}{n+1}),f(n)equiv {n^4-1equiv 0 {rm by} nnotequiv 0},$ and little Fermat



              so $,color{#c00}{nnotequiv -1}Rightarrow f(n)equiv 0$






              share|cite|improve this answer











              $endgroup$



              $!bmod 5!:, f(n)equiv n^2(overbrace{n!-!1}^{Large n - 6})+n!-!1, =, (color{#c00}{n!-!1}) (overbrace{n^2+1}^{Largecolor{#0a0}{n^2 - 4}})$



              therefore we conclude $ underbrace{ nnotequiv 0,-1}_{Large 5 nmid n(n+1)}Rightarrow color{#c00}{nequiv 1},$ or $!!!!underbrace{nequiv pm2}_{Large Rightarrow color{#0a0}{n^2 equiv 4} }!!!!$ so $,f(n)equiv 0$




              Or: $, (color{#c00}{n+1}),f(n)equiv {n^4-1equiv 0 {rm by} nnotequiv 0},$ and little Fermat



              so $,color{#c00}{nnotequiv -1}Rightarrow f(n)equiv 0$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 11 at 22:55

























              answered Jan 11 at 22:15









              Bill DubuqueBill Dubuque

              209k29191639




              209k29191639












              • $begingroup$
                We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
                $endgroup$
                – Bill Dubuque
                Jan 11 at 22:27




















              • $begingroup$
                We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
                $endgroup$
                – Bill Dubuque
                Jan 11 at 22:27


















              $begingroup$
              We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
              $endgroup$
              – Bill Dubuque
              Jan 11 at 22:27






              $begingroup$
              We used a balanced residue system $, nequiv 0,,pm1,pm2pmod{!5} $
              $endgroup$
              – Bill Dubuque
              Jan 11 at 22:27













              0












              $begingroup$

              $n^3 - 6n^2 + n-1 equiv n^3 - n^2 + n-1 pmod 5equiv$



              $(n^2 + 1)(n-1) equiv (n^2 -4)(n-1) pmod 5$



              $(n-2)(n+2)(n-1) pmod(5)$.



              If $n equiv 2,-2, 1 pmod 5$ then $n^3 - 5n^2 + n-1 equiv 0 pmod 5$ and $5|n^3 - 5n^2 + n-1$.



              As $5not mid n(n+1)$ and $5$ is prie $5not mid n$ and $5not mid n+1$ and $n not equiv 0pmod 5$ and $nnot equiv -1 pmod 5$.



              ${-2, -1, 0, 1, 2}$ is a complete residue class of $5$ so $nequiv -2,-1, 0, 1, $ or $2 pmod 5$ and since $nnot equiv 0,-1$ then $n equiv -2, 2$ or $1 pmod 5$. So we are done.



              ....



              Alternatively (now that we know where we want to go)



              $nnot mid n(n+1) implies$



              $n not mid 0$ or $-1pmod 5implies$



              $n mid 1,2,$ or $-2pmod 5implies$



              One of $n-1, n-2, n+2equiv 0 pmod 5implies$



              $(n-1)(n-2)(n+2)equiv 0 pmod 5 implies$



              $ n^3 -n^2 - 4n + 4equiv 0 pmod 5implies$



              $n^3 - 6n^2 + n - 1equiv 0 pmod 5 implies $



              $5|n^3 - 6n^2 + n-1$



              ....



              Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.



              $n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $n^3 - 6n^2 + n-1 equiv n^3 - n^2 + n-1 pmod 5equiv$



                $(n^2 + 1)(n-1) equiv (n^2 -4)(n-1) pmod 5$



                $(n-2)(n+2)(n-1) pmod(5)$.



                If $n equiv 2,-2, 1 pmod 5$ then $n^3 - 5n^2 + n-1 equiv 0 pmod 5$ and $5|n^3 - 5n^2 + n-1$.



                As $5not mid n(n+1)$ and $5$ is prie $5not mid n$ and $5not mid n+1$ and $n not equiv 0pmod 5$ and $nnot equiv -1 pmod 5$.



                ${-2, -1, 0, 1, 2}$ is a complete residue class of $5$ so $nequiv -2,-1, 0, 1, $ or $2 pmod 5$ and since $nnot equiv 0,-1$ then $n equiv -2, 2$ or $1 pmod 5$. So we are done.



                ....



                Alternatively (now that we know where we want to go)



                $nnot mid n(n+1) implies$



                $n not mid 0$ or $-1pmod 5implies$



                $n mid 1,2,$ or $-2pmod 5implies$



                One of $n-1, n-2, n+2equiv 0 pmod 5implies$



                $(n-1)(n-2)(n+2)equiv 0 pmod 5 implies$



                $ n^3 -n^2 - 4n + 4equiv 0 pmod 5implies$



                $n^3 - 6n^2 + n - 1equiv 0 pmod 5 implies $



                $5|n^3 - 6n^2 + n-1$



                ....



                Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.



                $n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $n^3 - 6n^2 + n-1 equiv n^3 - n^2 + n-1 pmod 5equiv$



                  $(n^2 + 1)(n-1) equiv (n^2 -4)(n-1) pmod 5$



                  $(n-2)(n+2)(n-1) pmod(5)$.



                  If $n equiv 2,-2, 1 pmod 5$ then $n^3 - 5n^2 + n-1 equiv 0 pmod 5$ and $5|n^3 - 5n^2 + n-1$.



                  As $5not mid n(n+1)$ and $5$ is prie $5not mid n$ and $5not mid n+1$ and $n not equiv 0pmod 5$ and $nnot equiv -1 pmod 5$.



                  ${-2, -1, 0, 1, 2}$ is a complete residue class of $5$ so $nequiv -2,-1, 0, 1, $ or $2 pmod 5$ and since $nnot equiv 0,-1$ then $n equiv -2, 2$ or $1 pmod 5$. So we are done.



                  ....



                  Alternatively (now that we know where we want to go)



                  $nnot mid n(n+1) implies$



                  $n not mid 0$ or $-1pmod 5implies$



                  $n mid 1,2,$ or $-2pmod 5implies$



                  One of $n-1, n-2, n+2equiv 0 pmod 5implies$



                  $(n-1)(n-2)(n+2)equiv 0 pmod 5 implies$



                  $ n^3 -n^2 - 4n + 4equiv 0 pmod 5implies$



                  $n^3 - 6n^2 + n - 1equiv 0 pmod 5 implies $



                  $5|n^3 - 6n^2 + n-1$



                  ....



                  Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.



                  $n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$






                  share|cite|improve this answer









                  $endgroup$



                  $n^3 - 6n^2 + n-1 equiv n^3 - n^2 + n-1 pmod 5equiv$



                  $(n^2 + 1)(n-1) equiv (n^2 -4)(n-1) pmod 5$



                  $(n-2)(n+2)(n-1) pmod(5)$.



                  If $n equiv 2,-2, 1 pmod 5$ then $n^3 - 5n^2 + n-1 equiv 0 pmod 5$ and $5|n^3 - 5n^2 + n-1$.



                  As $5not mid n(n+1)$ and $5$ is prie $5not mid n$ and $5not mid n+1$ and $n not equiv 0pmod 5$ and $nnot equiv -1 pmod 5$.



                  ${-2, -1, 0, 1, 2}$ is a complete residue class of $5$ so $nequiv -2,-1, 0, 1, $ or $2 pmod 5$ and since $nnot equiv 0,-1$ then $n equiv -2, 2$ or $1 pmod 5$. So we are done.



                  ....



                  Alternatively (now that we know where we want to go)



                  $nnot mid n(n+1) implies$



                  $n not mid 0$ or $-1pmod 5implies$



                  $n mid 1,2,$ or $-2pmod 5implies$



                  One of $n-1, n-2, n+2equiv 0 pmod 5implies$



                  $(n-1)(n-2)(n+2)equiv 0 pmod 5 implies$



                  $ n^3 -n^2 - 4n + 4equiv 0 pmod 5implies$



                  $n^3 - 6n^2 + n - 1equiv 0 pmod 5 implies $



                  $5|n^3 - 6n^2 + n-1$



                  ....



                  Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.



                  $n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 23:06









                  fleabloodfleablood

                  69.4k22685




                  69.4k22685






























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