Prove by mathematical induction that $n^4 − n^3 + n^2 − n$ is divisible by $2$ for all positive integers $n$.

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So I've done the base case with $n = 1$ and the general case of $k^4 − k^3 + k^2 − k = 2m$ but I can't seem to find a way to manipulate $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in a manner such that I can substitute in the $2m$. Any help?




elementary-number-theory induction divisibility




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  • Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
    – TheSimpliFire
    19 hours ago






  • 2




    Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
    – Simply Beautiful Art
    19 hours ago










  • Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
    – user485842
    19 hours ago






  • 1




    @SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
    – Ethan Bolker
    19 hours ago










  • Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
    – user485842
    19 hours ago














up vote
0
down vote

favorite












So I've done the base case with $n = 1$ and the general case of $k^4 − k^3 + k^2 − k = 2m$ but I can't seem to find a way to manipulate $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in a manner such that I can substitute in the $2m$. Any help?




elementary-number-theory induction divisibility




share|cite|improve this question





















  • Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
    – TheSimpliFire
    19 hours ago






  • 2




    Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
    – Simply Beautiful Art
    19 hours ago










  • Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
    – user485842
    19 hours ago






  • 1




    @SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
    – Ethan Bolker
    19 hours ago










  • Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
    – user485842
    19 hours ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite











So I've done the base case with $n = 1$ and the general case of $k^4 − k^3 + k^2 − k = 2m$ but I can't seem to find a way to manipulate $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in a manner such that I can substitute in the $2m$. Any help?




elementary-number-theory induction divisibility




share|cite|improve this question













So I've done the base case with $n = 1$ and the general case of $k^4 − k^3 + k^2 − k = 2m$ but I can't seem to find a way to manipulate $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in a manner such that I can substitute in the $2m$. Any help?





elementary-number-theory induction divisibility





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 19 hours ago









TheSimpliFire

9,18651651




9,18651651









asked 19 hours ago









user485842

426




426











  • Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
    – TheSimpliFire
    19 hours ago






  • 2




    Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
    – Simply Beautiful Art
    19 hours ago










  • Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
    – user485842
    19 hours ago






  • 1




    @SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
    – Ethan Bolker
    19 hours ago










  • Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
    – user485842
    19 hours ago
















  • Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
    – TheSimpliFire
    19 hours ago






  • 2




    Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
    – Simply Beautiful Art
    19 hours ago










  • Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
    – user485842
    19 hours ago






  • 1




    @SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
    – Ethan Bolker
    19 hours ago










  • Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
    – user485842
    19 hours ago















Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
– TheSimpliFire
19 hours ago




Alternatively, note that $$n^4-n^3+n^2-n=n(n^3-n^2+n-1)$$ and exactly one of $n$ and $n^3-n^2+n-1$ is even...
– TheSimpliFire
19 hours ago




2




2




Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
– Simply Beautiful Art
19 hours ago




Note: It's probably easier to prove this directly. If $n$ is even, you get even-even+even-even = even, and if $n$ is odd, you get odd-odd+odd-odd = even.
– Simply Beautiful Art
19 hours ago












Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
– user485842
19 hours ago




Hmmm. Yeah, I completely get it logically. J was just wondering if there was a way to somehow fit the number 2 into the proof of divisibility. Thanks.
– user485842
19 hours ago




1




1




@SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
– Ethan Bolker
19 hours ago




@SimplyBeautifulArt Indeed. I sigh whenever I see this kind of problem with instructions to use induction.
– Ethan Bolker
19 hours ago












Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
– user485842
19 hours ago




Thanks everyone. Our teacher just cleared it out. It doesn't require induction; the textbook isn't very reliable.
– user485842
19 hours ago










4 Answers
4






active

oldest

votes

















up vote
6
down vote



accepted










Hidden induction (almost directly):



$$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$



and since either $;n;$ or $;n-1;$ is even...






share|cite|improve this answer



















  • 2




    actually $n(n-1)(n^2+1)$
    – Henry
    19 hours ago










  • Though I think the answer solves the problem, but why is is inductive?
    – dEmigOd
    19 hours ago










  • @Henry Good catch, thanks. Edited.
    – DonAntonio
    19 hours ago











  • @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
    – DonAntonio
    19 hours ago






  • 2




    Long time no see, my friend !
    – Claude Leibovici
    19 hours ago

















up vote
2
down vote













Mod 2, $n$ takes only the two values 0 and 1 and so $n^kequiv n;[2]$. Hence



$$n^4-n^3+n^2-nequiv n-n+n-n=0;[2]$$






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    up vote
    2
    down vote













    Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.



    $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$



    $k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$



    $(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$



    $(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.



    As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.



    And we prove $3k^2 + 3k$ is even the exact same process.



    Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.



    Induction: Assume $3k^2 + 3k$ is even.



    Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.



    $3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$



    And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$






    share|cite|improve this answer




























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      Claim




      $n^4 − n^3 + n^2 − n$ is even for $n=1,2,cdots.$




      An Inductive Proof



      Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then beginalign* (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).endalign*



      Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.






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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        6
        down vote



        accepted










        Hidden induction (almost directly):



        $$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$



        and since either $;n;$ or $;n-1;$ is even...






        share|cite|improve this answer



















        • 2




          actually $n(n-1)(n^2+1)$
          – Henry
          19 hours ago










        • Though I think the answer solves the problem, but why is is inductive?
          – dEmigOd
          19 hours ago










        • @Henry Good catch, thanks. Edited.
          – DonAntonio
          19 hours ago











        • @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
          – DonAntonio
          19 hours ago






        • 2




          Long time no see, my friend !
          – Claude Leibovici
          19 hours ago














        up vote
        6
        down vote



        accepted










        Hidden induction (almost directly):



        $$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$



        and since either $;n;$ or $;n-1;$ is even...






        share|cite|improve this answer



















        • 2




          actually $n(n-1)(n^2+1)$
          – Henry
          19 hours ago










        • Though I think the answer solves the problem, but why is is inductive?
          – dEmigOd
          19 hours ago










        • @Henry Good catch, thanks. Edited.
          – DonAntonio
          19 hours ago











        • @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
          – DonAntonio
          19 hours ago






        • 2




          Long time no see, my friend !
          – Claude Leibovici
          19 hours ago












        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        Hidden induction (almost directly):



        $$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$



        and since either $;n;$ or $;n-1;$ is even...






        share|cite|improve this answer















        Hidden induction (almost directly):



        $$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$



        and since either $;n;$ or $;n-1;$ is even...







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 19 hours ago


























        answered 19 hours ago









        DonAntonio

        172k1483216




        172k1483216







        • 2




          actually $n(n-1)(n^2+1)$
          – Henry
          19 hours ago










        • Though I think the answer solves the problem, but why is is inductive?
          – dEmigOd
          19 hours ago










        • @Henry Good catch, thanks. Edited.
          – DonAntonio
          19 hours ago











        • @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
          – DonAntonio
          19 hours ago






        • 2




          Long time no see, my friend !
          – Claude Leibovici
          19 hours ago












        • 2




          actually $n(n-1)(n^2+1)$
          – Henry
          19 hours ago










        • Though I think the answer solves the problem, but why is is inductive?
          – dEmigOd
          19 hours ago










        • @Henry Good catch, thanks. Edited.
          – DonAntonio
          19 hours ago











        • @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
          – DonAntonio
          19 hours ago






        • 2




          Long time no see, my friend !
          – Claude Leibovici
          19 hours ago







        2




        2




        actually $n(n-1)(n^2+1)$
        – Henry
        19 hours ago




        actually $n(n-1)(n^2+1)$
        – Henry
        19 hours ago












        Though I think the answer solves the problem, but why is is inductive?
        – dEmigOd
        19 hours ago




        Though I think the answer solves the problem, but why is is inductive?
        – dEmigOd
        19 hours ago












        @Henry Good catch, thanks. Edited.
        – DonAntonio
        19 hours ago





        @Henry Good catch, thanks. Edited.
        – DonAntonio
        19 hours ago













        @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
        – DonAntonio
        19 hours ago




        @dEmigOd The "hidden" induction is, in this case, to prove that either $;n;$ or $;n-1;$ is even. It is basically ridiculously trivial, but formally it must be stated.
        – DonAntonio
        19 hours ago




        2




        2




        Long time no see, my friend !
        – Claude Leibovici
        19 hours ago




        Long time no see, my friend !
        – Claude Leibovici
        19 hours ago










        up vote
        2
        down vote













        Mod 2, $n$ takes only the two values 0 and 1 and so $n^kequiv n;[2]$. Hence



        $$n^4-n^3+n^2-nequiv n-n+n-n=0;[2]$$






        share|cite|improve this answer

























          up vote
          2
          down vote













          Mod 2, $n$ takes only the two values 0 and 1 and so $n^kequiv n;[2]$. Hence



          $$n^4-n^3+n^2-nequiv n-n+n-n=0;[2]$$






          share|cite|improve this answer























            up vote
            2
            down vote










            up vote
            2
            down vote









            Mod 2, $n$ takes only the two values 0 and 1 and so $n^kequiv n;[2]$. Hence



            $$n^4-n^3+n^2-nequiv n-n+n-n=0;[2]$$






            share|cite|improve this answer













            Mod 2, $n$ takes only the two values 0 and 1 and so $n^kequiv n;[2]$. Hence



            $$n^4-n^3+n^2-nequiv n-n+n-n=0;[2]$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 19 hours ago









            user296113

            6,464728




            6,464728




















                up vote
                2
                down vote













                Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.



                $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$



                $k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$



                $(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$



                $(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.



                As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.



                And we prove $3k^2 + 3k$ is even the exact same process.



                Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.



                Induction: Assume $3k^2 + 3k$ is even.



                Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.



                $3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$



                And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$






                share|cite|improve this answer

























                  up vote
                  2
                  down vote













                  Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.



                  $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$



                  $k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$



                  $(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$



                  $(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.



                  As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.



                  And we prove $3k^2 + 3k$ is even the exact same process.



                  Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.



                  Induction: Assume $3k^2 + 3k$ is even.



                  Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.



                  $3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$



                  And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$






                  share|cite|improve this answer























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.



                    $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$



                    $k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$



                    $(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$



                    $(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.



                    As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.



                    And we prove $3k^2 + 3k$ is even the exact same process.



                    Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.



                    Induction: Assume $3k^2 + 3k$ is even.



                    Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.



                    $3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$



                    And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$






                    share|cite|improve this answer













                    Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.



                    $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$



                    $k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$



                    $(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$



                    $(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.



                    As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.



                    And we prove $3k^2 + 3k$ is even the exact same process.



                    Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.



                    Induction: Assume $3k^2 + 3k$ is even.



                    Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.



                    $3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$



                    And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$







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                    answered 16 hours ago









                    fleablood

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                    60k22575




















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                        Claim




                        $n^4 − n^3 + n^2 − n$ is even for $n=1,2,cdots.$




                        An Inductive Proof



                        Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then beginalign* (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).endalign*



                        Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          Claim




                          $n^4 − n^3 + n^2 − n$ is even for $n=1,2,cdots.$




                          An Inductive Proof



                          Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then beginalign* (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).endalign*



                          Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Claim




                            $n^4 − n^3 + n^2 − n$ is even for $n=1,2,cdots.$




                            An Inductive Proof



                            Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then beginalign* (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).endalign*



                            Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.






                            share|cite|improve this answer













                            Claim




                            $n^4 − n^3 + n^2 − n$ is even for $n=1,2,cdots.$




                            An Inductive Proof



                            Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then beginalign* (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).endalign*



                            Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.







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                            answered 17 hours ago









                            mengdie1982

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