Prove that $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$ is natural number












0














Prove that for every $n in mathbb N$, $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$ is natural number.



I try in this form $frac{2n+2n^2+n^3}{6}=frac{n(n^2+2n+2)}{6}$ so I must show that $2| n(n^2+2n+2)$ and $3|n(n^2+2n+2)$ then I try that $n=2k$ or $n=2k+1$, $kin mathbb N$ to prove that if $n=2k$ it is trivial if $n=2k+1$ then $(2k+1)((2k+2)^2+1)$ then it is odd number multiply odd number so I did not prove that 2 divide that number, the same is for 3 do you have some idea?






discrete-mathematics modular-arithmetic divisibility







share|cite|improve this question


















  • 2




    It's not when $n=1$. Have you copied the question correctly?
    – Lord Shark the Unknown
    Nov 22 at 21:27










  • yes now i check but even for me it is so strange
    – Marko Škorić
    Nov 22 at 21:31










  • It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
    – fleablood
    Nov 22 at 21:32












  • You know that $n^3/6+n^2/2+n/3$ is an integer?
    – Lord Shark the Unknown
    Nov 22 at 21:32










  • See math.stackexchange.com/questions/710361/…
    – lab bhattacharjee
    Nov 23 at 4:09
















0














Prove that for every $n in mathbb N$, $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$ is natural number.



I try in this form $frac{2n+2n^2+n^3}{6}=frac{n(n^2+2n+2)}{6}$ so I must show that $2| n(n^2+2n+2)$ and $3|n(n^2+2n+2)$ then I try that $n=2k$ or $n=2k+1$, $kin mathbb N$ to prove that if $n=2k$ it is trivial if $n=2k+1$ then $(2k+1)((2k+2)^2+1)$ then it is odd number multiply odd number so I did not prove that 2 divide that number, the same is for 3 do you have some idea?






discrete-mathematics modular-arithmetic divisibility







share|cite|improve this question


















  • 2




    It's not when $n=1$. Have you copied the question correctly?
    – Lord Shark the Unknown
    Nov 22 at 21:27










  • yes now i check but even for me it is so strange
    – Marko Škorić
    Nov 22 at 21:31










  • It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
    – fleablood
    Nov 22 at 21:32












  • You know that $n^3/6+n^2/2+n/3$ is an integer?
    – Lord Shark the Unknown
    Nov 22 at 21:32










  • See math.stackexchange.com/questions/710361/…
    – lab bhattacharjee
    Nov 23 at 4:09














0












0








0







Prove that for every $n in mathbb N$, $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$ is natural number.



I try in this form $frac{2n+2n^2+n^3}{6}=frac{n(n^2+2n+2)}{6}$ so I must show that $2| n(n^2+2n+2)$ and $3|n(n^2+2n+2)$ then I try that $n=2k$ or $n=2k+1$, $kin mathbb N$ to prove that if $n=2k$ it is trivial if $n=2k+1$ then $(2k+1)((2k+2)^2+1)$ then it is odd number multiply odd number so I did not prove that 2 divide that number, the same is for 3 do you have some idea?






discrete-mathematics modular-arithmetic divisibility







share|cite|improve this question













Prove that for every $n in mathbb N$, $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$ is natural number.



I try in this form $frac{2n+2n^2+n^3}{6}=frac{n(n^2+2n+2)}{6}$ so I must show that $2| n(n^2+2n+2)$ and $3|n(n^2+2n+2)$ then I try that $n=2k$ or $n=2k+1$, $kin mathbb N$ to prove that if $n=2k$ it is trivial if $n=2k+1$ then $(2k+1)((2k+2)^2+1)$ then it is odd number multiply odd number so I did not prove that 2 divide that number, the same is for 3 do you have some idea?






discrete-mathematics modular-arithmetic divisibility




discrete-mathematics modular-arithmetic divisibility






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 21:24









Marko Škorić

70310




70310








  • 2




    It's not when $n=1$. Have you copied the question correctly?
    – Lord Shark the Unknown
    Nov 22 at 21:27










  • yes now i check but even for me it is so strange
    – Marko Škorić
    Nov 22 at 21:31










  • It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
    – fleablood
    Nov 22 at 21:32












  • You know that $n^3/6+n^2/2+n/3$ is an integer?
    – Lord Shark the Unknown
    Nov 22 at 21:32










  • See math.stackexchange.com/questions/710361/…
    – lab bhattacharjee
    Nov 23 at 4:09














  • 2




    It's not when $n=1$. Have you copied the question correctly?
    – Lord Shark the Unknown
    Nov 22 at 21:27










  • yes now i check but even for me it is so strange
    – Marko Škorić
    Nov 22 at 21:31










  • It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
    – fleablood
    Nov 22 at 21:32












  • You know that $n^3/6+n^2/2+n/3$ is an integer?
    – Lord Shark the Unknown
    Nov 22 at 21:32










  • See math.stackexchange.com/questions/710361/…
    – lab bhattacharjee
    Nov 23 at 4:09








2




2




It's not when $n=1$. Have you copied the question correctly?
– Lord Shark the Unknown
Nov 22 at 21:27




It's not when $n=1$. Have you copied the question correctly?
– Lord Shark the Unknown
Nov 22 at 21:27












yes now i check but even for me it is so strange
– Marko Škorić
Nov 22 at 21:31




yes now i check but even for me it is so strange
– Marko Škorić
Nov 22 at 21:31












It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
– fleablood
Nov 22 at 21:32






It seems it won't be true if n is odd. If n is 3 for example it is $1+3+frac 92$
– fleablood
Nov 22 at 21:32














You know that $n^3/6+n^2/2+n/3$ is an integer?
– Lord Shark the Unknown
Nov 22 at 21:32




You know that $n^3/6+n^2/2+n/3$ is an integer?
– Lord Shark the Unknown
Nov 22 at 21:32












See math.stackexchange.com/questions/710361/…
– lab bhattacharjee
Nov 23 at 4:09




See math.stackexchange.com/questions/710361/…
– lab bhattacharjee
Nov 23 at 4:09










4 Answers
4






active

oldest

votes


















1














By below $,6mid nf(n)!=! n(n^2!+3n+2) $ by $ 6mid f(1)!=6,,$ and $,3mid f(-1)!=!0$




Theorem $ forall n!: 6mid nf(n)! iff! 6mid f(1), 3mid f(-1) $ for a polynomial $f(x)$ with integer coef's



Proof $ $ It is true $iff nf(n),$ has roots $,nequiv 0,1pmod{! 2}$ and $,nequiv 0,pm1 pmod{!3}$



But $,0,$ is always a root, and $,1,$ is a root $!iff!!! underbrace{6mid f(1)}_{large 2,3 mid f(1) }!$ and $,-1$ is a root $iff!3mid f(-1)$






share|cite|improve this answer































    3














    It is a typo because $$frac{n}{3}+frac{n^2}{2}+frac{n^3}{6}={n(n+1)(n+2)over 6}$$ is natural number for all $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$






    share|cite|improve this answer





















    • yep, I suspect the same
      – Masacroso
      Nov 22 at 21:35










    • Just requires some manipulation in numerator or denominators........
      – Mostafa Ayaz
      Nov 22 at 21:36



















    2














    $m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.






    share|cite|improve this answer





























      0














      Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, as a consequence of the properties of the forward difference operator $delta:p(x)mapsto p(x+1)-p(x)$. Since
      $$ frac{n}{3}+frac{n^2}{color{red}{2}}+frac{n^3}{6} $$
      takes integer values at ${0,1,2,3}$, it takes integer values at any $ninmathbb{N}$, and these values are clearly non-negative.






      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









        1














        By below $,6mid nf(n)!=! n(n^2!+3n+2) $ by $ 6mid f(1)!=6,,$ and $,3mid f(-1)!=!0$




        Theorem $ forall n!: 6mid nf(n)! iff! 6mid f(1), 3mid f(-1) $ for a polynomial $f(x)$ with integer coef's



        Proof $ $ It is true $iff nf(n),$ has roots $,nequiv 0,1pmod{! 2}$ and $,nequiv 0,pm1 pmod{!3}$



        But $,0,$ is always a root, and $,1,$ is a root $!iff!!! underbrace{6mid f(1)}_{large 2,3 mid f(1) }!$ and $,-1$ is a root $iff!3mid f(-1)$






        share|cite|improve this answer




























          1














          By below $,6mid nf(n)!=! n(n^2!+3n+2) $ by $ 6mid f(1)!=6,,$ and $,3mid f(-1)!=!0$




          Theorem $ forall n!: 6mid nf(n)! iff! 6mid f(1), 3mid f(-1) $ for a polynomial $f(x)$ with integer coef's



          Proof $ $ It is true $iff nf(n),$ has roots $,nequiv 0,1pmod{! 2}$ and $,nequiv 0,pm1 pmod{!3}$



          But $,0,$ is always a root, and $,1,$ is a root $!iff!!! underbrace{6mid f(1)}_{large 2,3 mid f(1) }!$ and $,-1$ is a root $iff!3mid f(-1)$






          share|cite|improve this answer


























            1












            1








            1






            By below $,6mid nf(n)!=! n(n^2!+3n+2) $ by $ 6mid f(1)!=6,,$ and $,3mid f(-1)!=!0$




            Theorem $ forall n!: 6mid nf(n)! iff! 6mid f(1), 3mid f(-1) $ for a polynomial $f(x)$ with integer coef's



            Proof $ $ It is true $iff nf(n),$ has roots $,nequiv 0,1pmod{! 2}$ and $,nequiv 0,pm1 pmod{!3}$



            But $,0,$ is always a root, and $,1,$ is a root $!iff!!! underbrace{6mid f(1)}_{large 2,3 mid f(1) }!$ and $,-1$ is a root $iff!3mid f(-1)$






            share|cite|improve this answer














            By below $,6mid nf(n)!=! n(n^2!+3n+2) $ by $ 6mid f(1)!=6,,$ and $,3mid f(-1)!=!0$




            Theorem $ forall n!: 6mid nf(n)! iff! 6mid f(1), 3mid f(-1) $ for a polynomial $f(x)$ with integer coef's



            Proof $ $ It is true $iff nf(n),$ has roots $,nequiv 0,1pmod{! 2}$ and $,nequiv 0,pm1 pmod{!3}$



            But $,0,$ is always a root, and $,1,$ is a root $!iff!!! underbrace{6mid f(1)}_{large 2,3 mid f(1) }!$ and $,-1$ is a root $iff!3mid f(-1)$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 22 at 23:21

























            answered Nov 22 at 22:17









            Bill Dubuque

            208k29190626




            208k29190626























                3














                It is a typo because $$frac{n}{3}+frac{n^2}{2}+frac{n^3}{6}={n(n+1)(n+2)over 6}$$ is natural number for all $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$






                share|cite|improve this answer





















                • yep, I suspect the same
                  – Masacroso
                  Nov 22 at 21:35










                • Just requires some manipulation in numerator or denominators........
                  – Mostafa Ayaz
                  Nov 22 at 21:36
















                3














                It is a typo because $$frac{n}{3}+frac{n^2}{2}+frac{n^3}{6}={n(n+1)(n+2)over 6}$$ is natural number for all $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$






                share|cite|improve this answer





















                • yep, I suspect the same
                  – Masacroso
                  Nov 22 at 21:35










                • Just requires some manipulation in numerator or denominators........
                  – Mostafa Ayaz
                  Nov 22 at 21:36














                3












                3








                3






                It is a typo because $$frac{n}{3}+frac{n^2}{2}+frac{n^3}{6}={n(n+1)(n+2)over 6}$$ is natural number for all $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$






                share|cite|improve this answer












                It is a typo because $$frac{n}{3}+frac{n^2}{2}+frac{n^3}{6}={n(n+1)(n+2)over 6}$$ is natural number for all $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 21:33









                Mostafa Ayaz

                13.7k3836




                13.7k3836












                • yep, I suspect the same
                  – Masacroso
                  Nov 22 at 21:35










                • Just requires some manipulation in numerator or denominators........
                  – Mostafa Ayaz
                  Nov 22 at 21:36


















                • yep, I suspect the same
                  – Masacroso
                  Nov 22 at 21:35










                • Just requires some manipulation in numerator or denominators........
                  – Mostafa Ayaz
                  Nov 22 at 21:36
















                yep, I suspect the same
                – Masacroso
                Nov 22 at 21:35




                yep, I suspect the same
                – Masacroso
                Nov 22 at 21:35












                Just requires some manipulation in numerator or denominators........
                – Mostafa Ayaz
                Nov 22 at 21:36




                Just requires some manipulation in numerator or denominators........
                – Mostafa Ayaz
                Nov 22 at 21:36











                2














                $m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.






                share|cite|improve this answer


























                  2














                  $m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    $m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.






                    share|cite|improve this answer












                    $m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 21:32









                    John_Wick

                    1,356111




                    1,356111























                        0














                        Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, as a consequence of the properties of the forward difference operator $delta:p(x)mapsto p(x+1)-p(x)$. Since
                        $$ frac{n}{3}+frac{n^2}{color{red}{2}}+frac{n^3}{6} $$
                        takes integer values at ${0,1,2,3}$, it takes integer values at any $ninmathbb{N}$, and these values are clearly non-negative.






                        share|cite|improve this answer


























                          0














                          Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, as a consequence of the properties of the forward difference operator $delta:p(x)mapsto p(x+1)-p(x)$. Since
                          $$ frac{n}{3}+frac{n^2}{color{red}{2}}+frac{n^3}{6} $$
                          takes integer values at ${0,1,2,3}$, it takes integer values at any $ninmathbb{N}$, and these values are clearly non-negative.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, as a consequence of the properties of the forward difference operator $delta:p(x)mapsto p(x+1)-p(x)$. Since
                            $$ frac{n}{3}+frac{n^2}{color{red}{2}}+frac{n^3}{6} $$
                            takes integer values at ${0,1,2,3}$, it takes integer values at any $ninmathbb{N}$, and these values are clearly non-negative.






                            share|cite|improve this answer












                            Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, as a consequence of the properties of the forward difference operator $delta:p(x)mapsto p(x+1)-p(x)$. Since
                            $$ frac{n}{3}+frac{n^2}{color{red}{2}}+frac{n^3}{6} $$
                            takes integer values at ${0,1,2,3}$, it takes integer values at any $ninmathbb{N}$, and these values are clearly non-negative.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 22:32









                            Jack D'Aurizio

                            286k33279655




                            286k33279655






























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                                Cáceres

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                                Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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