Find $n$ such that polynomial is divisible












1












$begingroup$


Find $n in N$, such that:
$$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$



If I let $Q(x) = x^2 + x + 1$, I have that $x^3 equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
I suppose this could be done with complex zeroes, but is there a faster way ?






polynomials divisibility







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Find $n in N$, such that:
    $$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$



    If I let $Q(x) = x^2 + x + 1$, I have that $x^3 equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
    But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
    I suppose this could be done with complex zeroes, but is there a faster way ?






    polynomials divisibility







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Find $n in N$, such that:
      $$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$



      If I let $Q(x) = x^2 + x + 1$, I have that $x^3 equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
      But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
      I suppose this could be done with complex zeroes, but is there a faster way ?






      polynomials divisibility







      share|cite|improve this question











      $endgroup$




      Find $n in N$, such that:
      $$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$



      If I let $Q(x) = x^2 + x + 1$, I have that $x^3 equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
      But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
      I suppose this could be done with complex zeroes, but is there a faster way ?






      polynomials divisibility




      polynomials divisibility






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 24 at 15:40

























      asked Jan 24 at 15:35







      user626177





























          4 Answers
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          1












          $begingroup$

          Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.



          Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
          $$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
          hence
          $$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
          for all $kinmathbb{N}$, and similarly
          $$(x+1)^k=(-1)^kx^{2k}.$$



          Then we get
          $$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
          so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
          and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
          $$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
          which can never be zero.



          Hence $Q(x)^2$ divides $P(x)$ iff $nequiv 4pmod{6}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How come that $Q(x)$ divides derivative of $P(x)$ ?
            $endgroup$
            – user626177
            Jan 24 at 17:15










          • $begingroup$
            @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
            $endgroup$
            – Jose Brox
            Jan 24 at 17:33












          • $begingroup$
            I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
            $endgroup$
            – user626177
            Jan 24 at 17:39










          • $begingroup$
            @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
            $endgroup$
            – Jose Brox
            Jan 24 at 19:28










          • $begingroup$
            I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
            $endgroup$
            – user626177
            Jan 24 at 19:45



















          0












          $begingroup$

          Since the LHS is $(x-w)^{2}(x-w^{2})^{2}$ where $w = e^{2pi i /3}$, we need to find $n$ such that $P(w) = P(w^{2}) = 0$ and $P'(w) = P'(w^{2}) = 0$. You may use the identity $w^{2} + w + 1 = 0$ for this.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I'm not quite familiar with that type of complex exponentation
            $endgroup$
            – user626177
            Jan 24 at 16:20



















          0












          $begingroup$

          Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $omega$ is a root of $x^2 + x + 1$, we want
          $omega$ to be a root of $P_n(x)$ of multiplicity $ge 2$,
          which means both $P_n(omega) = omega^n + (omega+1)^n + 1 = 0$ and
          $P_n'(omega) = n omega^{n-1} + n (omega+1)^{n-1} = 0$. This last
          simplifies to $$left(1 + frac{1}{omega}right)^{n-1} = -1$$
          Now if $omega^2 + omega+1 = 0$ we have $omega + 1 + 1/omega = 0$, i.e. $1+1/omega = -omega$. Since the powers of $omega$ repeat $1, omega, omega^2$, we see that this is true if and only if $n-1 equiv 3 mod 6$, i.e. $n equiv 4 mod 6$. For such $n$ we also have
          $P_n(omega) = omega^n + (-omega^2)^n + 1 = omega + omega^2 + 1 = 0$, so both conditions are true.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint $ $ A double root is a root of $P,P'$ so also $, frac{1}{n}(x!+!1)P'!-!P = x^{large n-1}!-!1,Rightarrow, 3mid n!-!1,,$ by $,x^{large 3}!=1$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              See also this answer.
              $endgroup$
              – Bill Dubuque
              Jan 24 at 17:44










            • $begingroup$
              It seems like you are familiar with every generalization
              $endgroup$
              – user626177
              Jan 24 at 17:54






            • 1




              $begingroup$
              @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
              $endgroup$
              – Bill Dubuque
              Jan 24 at 17:56













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            4 Answers
            4






            active

            oldest

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            4 Answers
            4






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            1












            $begingroup$

            Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.



            Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
            $$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
            hence
            $$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
            for all $kinmathbb{N}$, and similarly
            $$(x+1)^k=(-1)^kx^{2k}.$$



            Then we get
            $$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
            so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
            and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
            $$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
            which can never be zero.



            Hence $Q(x)^2$ divides $P(x)$ iff $nequiv 4pmod{6}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How come that $Q(x)$ divides derivative of $P(x)$ ?
              $endgroup$
              – user626177
              Jan 24 at 17:15










            • $begingroup$
              @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
              $endgroup$
              – Jose Brox
              Jan 24 at 17:33












            • $begingroup$
              I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
              $endgroup$
              – user626177
              Jan 24 at 17:39










            • $begingroup$
              @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
              $endgroup$
              – Jose Brox
              Jan 24 at 19:28










            • $begingroup$
              I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
              $endgroup$
              – user626177
              Jan 24 at 19:45
















            1












            $begingroup$

            Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.



            Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
            $$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
            hence
            $$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
            for all $kinmathbb{N}$, and similarly
            $$(x+1)^k=(-1)^kx^{2k}.$$



            Then we get
            $$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
            so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
            and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
            $$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
            which can never be zero.



            Hence $Q(x)^2$ divides $P(x)$ iff $nequiv 4pmod{6}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How come that $Q(x)$ divides derivative of $P(x)$ ?
              $endgroup$
              – user626177
              Jan 24 at 17:15










            • $begingroup$
              @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
              $endgroup$
              – Jose Brox
              Jan 24 at 17:33












            • $begingroup$
              I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
              $endgroup$
              – user626177
              Jan 24 at 17:39










            • $begingroup$
              @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
              $endgroup$
              – Jose Brox
              Jan 24 at 19:28










            • $begingroup$
              I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
              $endgroup$
              – user626177
              Jan 24 at 19:45














            1












            1








            1





            $begingroup$

            Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.



            Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
            $$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
            hence
            $$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
            for all $kinmathbb{N}$, and similarly
            $$(x+1)^k=(-1)^kx^{2k}.$$



            Then we get
            $$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
            so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
            and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
            $$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
            which can never be zero.



            Hence $Q(x)^2$ divides $P(x)$ iff $nequiv 4pmod{6}$.






            share|cite|improve this answer









            $endgroup$



            Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.



            Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
            $$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
            hence
            $$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
            for all $kinmathbb{N}$, and similarly
            $$(x+1)^k=(-1)^kx^{2k}.$$



            Then we get
            $$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
            so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
            and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
            $$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
            which can never be zero.



            Hence $Q(x)^2$ divides $P(x)$ iff $nequiv 4pmod{6}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 17:11









            Jose BroxJose Brox

            3,15711128




            3,15711128












            • $begingroup$
              How come that $Q(x)$ divides derivative of $P(x)$ ?
              $endgroup$
              – user626177
              Jan 24 at 17:15










            • $begingroup$
              @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
              $endgroup$
              – Jose Brox
              Jan 24 at 17:33












            • $begingroup$
              I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
              $endgroup$
              – user626177
              Jan 24 at 17:39










            • $begingroup$
              @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
              $endgroup$
              – Jose Brox
              Jan 24 at 19:28










            • $begingroup$
              I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
              $endgroup$
              – user626177
              Jan 24 at 19:45


















            • $begingroup$
              How come that $Q(x)$ divides derivative of $P(x)$ ?
              $endgroup$
              – user626177
              Jan 24 at 17:15










            • $begingroup$
              @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
              $endgroup$
              – Jose Brox
              Jan 24 at 17:33












            • $begingroup$
              I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
              $endgroup$
              – user626177
              Jan 24 at 17:39










            • $begingroup$
              @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
              $endgroup$
              – Jose Brox
              Jan 24 at 19:28










            • $begingroup$
              I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
              $endgroup$
              – user626177
              Jan 24 at 19:45
















            $begingroup$
            How come that $Q(x)$ divides derivative of $P(x)$ ?
            $endgroup$
            – user626177
            Jan 24 at 17:15




            $begingroup$
            How come that $Q(x)$ divides derivative of $P(x)$ ?
            $endgroup$
            – user626177
            Jan 24 at 17:15












            $begingroup$
            @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
            $endgroup$
            – Jose Brox
            Jan 24 at 17:33






            $begingroup$
            @someone Since $Q(x)^2$ divides $P(x)$, we have $P(x)=Q(x)^2R(x)$ for some polynomial $R$. Then by the product rule we get $P'(x)=2Q(x)Q'(x)R(x)+Q(x)^2R'(x)=Q(x)(2Q'(x)R(x)+Q(x)R'(x))$, so $Q$ divides $P'$.
            $endgroup$
            – Jose Brox
            Jan 24 at 17:33














            $begingroup$
            I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
            $endgroup$
            – user626177
            Jan 24 at 17:39




            $begingroup$
            I thought you somehow derived that from this theorem: "If a is a root of multiplicity k of a polynomial, then it is a root of multiplicity k – 1 of its derivative". (This is completely different thing, right ?) Thanks!
            $endgroup$
            – user626177
            Jan 24 at 17:39












            $begingroup$
            @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
            $endgroup$
            – Jose Brox
            Jan 24 at 19:28




            $begingroup$
            @someone No, it is the same thing, for the particular case $k=2$! The proof of the theorem is the same, just considering $k$ instead of $2$. I'll let it to you as a practising exercise!
            $endgroup$
            – Jose Brox
            Jan 24 at 19:28












            $begingroup$
            I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
            $endgroup$
            – user626177
            Jan 24 at 19:45




            $begingroup$
            I was somehow thinking that root of a polynomial is $a in C$ and not a polynomial
            $endgroup$
            – user626177
            Jan 24 at 19:45











            0












            $begingroup$

            Since the LHS is $(x-w)^{2}(x-w^{2})^{2}$ where $w = e^{2pi i /3}$, we need to find $n$ such that $P(w) = P(w^{2}) = 0$ and $P'(w) = P'(w^{2}) = 0$. You may use the identity $w^{2} + w + 1 = 0$ for this.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm not quite familiar with that type of complex exponentation
              $endgroup$
              – user626177
              Jan 24 at 16:20
















            0












            $begingroup$

            Since the LHS is $(x-w)^{2}(x-w^{2})^{2}$ where $w = e^{2pi i /3}$, we need to find $n$ such that $P(w) = P(w^{2}) = 0$ and $P'(w) = P'(w^{2}) = 0$. You may use the identity $w^{2} + w + 1 = 0$ for this.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I'm not quite familiar with that type of complex exponentation
              $endgroup$
              – user626177
              Jan 24 at 16:20














            0












            0








            0





            $begingroup$

            Since the LHS is $(x-w)^{2}(x-w^{2})^{2}$ where $w = e^{2pi i /3}$, we need to find $n$ such that $P(w) = P(w^{2}) = 0$ and $P'(w) = P'(w^{2}) = 0$. You may use the identity $w^{2} + w + 1 = 0$ for this.






            share|cite|improve this answer









            $endgroup$



            Since the LHS is $(x-w)^{2}(x-w^{2})^{2}$ where $w = e^{2pi i /3}$, we need to find $n$ such that $P(w) = P(w^{2}) = 0$ and $P'(w) = P'(w^{2}) = 0$. You may use the identity $w^{2} + w + 1 = 0$ for this.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 15:58









            Seewoo LeeSeewoo Lee

            7,037927




            7,037927












            • $begingroup$
              I'm not quite familiar with that type of complex exponentation
              $endgroup$
              – user626177
              Jan 24 at 16:20


















            • $begingroup$
              I'm not quite familiar with that type of complex exponentation
              $endgroup$
              – user626177
              Jan 24 at 16:20
















            $begingroup$
            I'm not quite familiar with that type of complex exponentation
            $endgroup$
            – user626177
            Jan 24 at 16:20




            $begingroup$
            I'm not quite familiar with that type of complex exponentation
            $endgroup$
            – user626177
            Jan 24 at 16:20











            0












            $begingroup$

            Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $omega$ is a root of $x^2 + x + 1$, we want
            $omega$ to be a root of $P_n(x)$ of multiplicity $ge 2$,
            which means both $P_n(omega) = omega^n + (omega+1)^n + 1 = 0$ and
            $P_n'(omega) = n omega^{n-1} + n (omega+1)^{n-1} = 0$. This last
            simplifies to $$left(1 + frac{1}{omega}right)^{n-1} = -1$$
            Now if $omega^2 + omega+1 = 0$ we have $omega + 1 + 1/omega = 0$, i.e. $1+1/omega = -omega$. Since the powers of $omega$ repeat $1, omega, omega^2$, we see that this is true if and only if $n-1 equiv 3 mod 6$, i.e. $n equiv 4 mod 6$. For such $n$ we also have
            $P_n(omega) = omega^n + (-omega^2)^n + 1 = omega + omega^2 + 1 = 0$, so both conditions are true.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $omega$ is a root of $x^2 + x + 1$, we want
              $omega$ to be a root of $P_n(x)$ of multiplicity $ge 2$,
              which means both $P_n(omega) = omega^n + (omega+1)^n + 1 = 0$ and
              $P_n'(omega) = n omega^{n-1} + n (omega+1)^{n-1} = 0$. This last
              simplifies to $$left(1 + frac{1}{omega}right)^{n-1} = -1$$
              Now if $omega^2 + omega+1 = 0$ we have $omega + 1 + 1/omega = 0$, i.e. $1+1/omega = -omega$. Since the powers of $omega$ repeat $1, omega, omega^2$, we see that this is true if and only if $n-1 equiv 3 mod 6$, i.e. $n equiv 4 mod 6$. For such $n$ we also have
              $P_n(omega) = omega^n + (-omega^2)^n + 1 = omega + omega^2 + 1 = 0$, so both conditions are true.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $omega$ is a root of $x^2 + x + 1$, we want
                $omega$ to be a root of $P_n(x)$ of multiplicity $ge 2$,
                which means both $P_n(omega) = omega^n + (omega+1)^n + 1 = 0$ and
                $P_n'(omega) = n omega^{n-1} + n (omega+1)^{n-1} = 0$. This last
                simplifies to $$left(1 + frac{1}{omega}right)^{n-1} = -1$$
                Now if $omega^2 + omega+1 = 0$ we have $omega + 1 + 1/omega = 0$, i.e. $1+1/omega = -omega$. Since the powers of $omega$ repeat $1, omega, omega^2$, we see that this is true if and only if $n-1 equiv 3 mod 6$, i.e. $n equiv 4 mod 6$. For such $n$ we also have
                $P_n(omega) = omega^n + (-omega^2)^n + 1 = omega + omega^2 + 1 = 0$, so both conditions are true.






                share|cite|improve this answer









                $endgroup$



                Let's call your polynomial $P_n(x)$, to indicate the dependence on $n$. If $omega$ is a root of $x^2 + x + 1$, we want
                $omega$ to be a root of $P_n(x)$ of multiplicity $ge 2$,
                which means both $P_n(omega) = omega^n + (omega+1)^n + 1 = 0$ and
                $P_n'(omega) = n omega^{n-1} + n (omega+1)^{n-1} = 0$. This last
                simplifies to $$left(1 + frac{1}{omega}right)^{n-1} = -1$$
                Now if $omega^2 + omega+1 = 0$ we have $omega + 1 + 1/omega = 0$, i.e. $1+1/omega = -omega$. Since the powers of $omega$ repeat $1, omega, omega^2$, we see that this is true if and only if $n-1 equiv 3 mod 6$, i.e. $n equiv 4 mod 6$. For such $n$ we also have
                $P_n(omega) = omega^n + (-omega^2)^n + 1 = omega + omega^2 + 1 = 0$, so both conditions are true.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 24 at 16:40









                Robert IsraelRobert Israel

                326k23215469




                326k23215469























                    0












                    $begingroup$

                    Hint $ $ A double root is a root of $P,P'$ so also $, frac{1}{n}(x!+!1)P'!-!P = x^{large n-1}!-!1,Rightarrow, 3mid n!-!1,,$ by $,x^{large 3}!=1$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      See also this answer.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:44










                    • $begingroup$
                      It seems like you are familiar with every generalization
                      $endgroup$
                      – user626177
                      Jan 24 at 17:54






                    • 1




                      $begingroup$
                      @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:56


















                    0












                    $begingroup$

                    Hint $ $ A double root is a root of $P,P'$ so also $, frac{1}{n}(x!+!1)P'!-!P = x^{large n-1}!-!1,Rightarrow, 3mid n!-!1,,$ by $,x^{large 3}!=1$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      See also this answer.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:44










                    • $begingroup$
                      It seems like you are familiar with every generalization
                      $endgroup$
                      – user626177
                      Jan 24 at 17:54






                    • 1




                      $begingroup$
                      @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:56
















                    0












                    0








                    0





                    $begingroup$

                    Hint $ $ A double root is a root of $P,P'$ so also $, frac{1}{n}(x!+!1)P'!-!P = x^{large n-1}!-!1,Rightarrow, 3mid n!-!1,,$ by $,x^{large 3}!=1$






                    share|cite|improve this answer











                    $endgroup$



                    Hint $ $ A double root is a root of $P,P'$ so also $, frac{1}{n}(x!+!1)P'!-!P = x^{large n-1}!-!1,Rightarrow, 3mid n!-!1,,$ by $,x^{large 3}!=1$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 24 at 17:48

























                    answered Jan 24 at 17:42









                    Bill DubuqueBill Dubuque

                    212k29195650




                    212k29195650












                    • $begingroup$
                      See also this answer.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:44










                    • $begingroup$
                      It seems like you are familiar with every generalization
                      $endgroup$
                      – user626177
                      Jan 24 at 17:54






                    • 1




                      $begingroup$
                      @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:56




















                    • $begingroup$
                      See also this answer.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:44










                    • $begingroup$
                      It seems like you are familiar with every generalization
                      $endgroup$
                      – user626177
                      Jan 24 at 17:54






                    • 1




                      $begingroup$
                      @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                      $endgroup$
                      – Bill Dubuque
                      Jan 24 at 17:56


















                    $begingroup$
                    See also this answer.
                    $endgroup$
                    – Bill Dubuque
                    Jan 24 at 17:44




                    $begingroup$
                    See also this answer.
                    $endgroup$
                    – Bill Dubuque
                    Jan 24 at 17:44












                    $begingroup$
                    It seems like you are familiar with every generalization
                    $endgroup$
                    – user626177
                    Jan 24 at 17:54




                    $begingroup$
                    It seems like you are familiar with every generalization
                    $endgroup$
                    – user626177
                    Jan 24 at 17:54




                    1




                    1




                    $begingroup$
                    @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                    $endgroup$
                    – Bill Dubuque
                    Jan 24 at 17:56






                    $begingroup$
                    @someone Modular arithmetic makes it easy - as I explain in the linked answer. The above is essentially an extension of that to deal with double factors.
                    $endgroup$
                    – Bill Dubuque
                    Jan 24 at 17:56




















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