$f^n+g^n=h^2implies f,g,h$ all constant












4












$begingroup$


Let $k$ be a field with char$(k)=0$ and suppose for $f,g,hin k[x]$ having gcd$(f,g,h)=1$ and $ninmathbb{N}_{geqslant4}$ it holds that $f^n+g^n=h^2$. I want to show that $f,g,h$ are all constant. Of course, we go the Mason Stothers-way. Quick note: by gcd $1$ and the equation, we have that $f,g,h$ are pairwise co-prime.



For the sake of contradiction, suppose that $f,g,h$ are not all constant. By the other conditions listed above, we can apply Mason-Stothers like so:
$$max(deg f^n,deg g^n,deg h^2)leqslantdeg(text{rad}(fgh))-1.$$
Now let's look at two cases:



1. $f^n$ has maximal degree (and is non-constant).



2. $h^2$ has maximal degree.



A contradiction should appear in both cases. Let's begin with 1.:
Since $f^n$ has max degree and the polynomials are pairwise co-prime, we can say $$begin{align*}4leqslantdeg(f^n)=ndeg(f)&<deg({text{rad}(fgh)})\
&= deg(text{rad}(f)text{rad}(g)text{rad}(h))\
&=deg(text{rad}(f))+deg(text{rad}(g))+deg(text{rad}(h))\
&leqslant3deg(f).
end{align*}$$
Now since $n>3$, this is an immediate contradiction.



Thanks for all the fast help!






number-theory polynomials field-theory radicals







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$endgroup$

















    4












    $begingroup$


    Let $k$ be a field with char$(k)=0$ and suppose for $f,g,hin k[x]$ having gcd$(f,g,h)=1$ and $ninmathbb{N}_{geqslant4}$ it holds that $f^n+g^n=h^2$. I want to show that $f,g,h$ are all constant. Of course, we go the Mason Stothers-way. Quick note: by gcd $1$ and the equation, we have that $f,g,h$ are pairwise co-prime.



    For the sake of contradiction, suppose that $f,g,h$ are not all constant. By the other conditions listed above, we can apply Mason-Stothers like so:
    $$max(deg f^n,deg g^n,deg h^2)leqslantdeg(text{rad}(fgh))-1.$$
    Now let's look at two cases:



    1. $f^n$ has maximal degree (and is non-constant).



    2. $h^2$ has maximal degree.



    A contradiction should appear in both cases. Let's begin with 1.:
    Since $f^n$ has max degree and the polynomials are pairwise co-prime, we can say $$begin{align*}4leqslantdeg(f^n)=ndeg(f)&<deg({text{rad}(fgh)})\
    &= deg(text{rad}(f)text{rad}(g)text{rad}(h))\
    &=deg(text{rad}(f))+deg(text{rad}(g))+deg(text{rad}(h))\
    &leqslant3deg(f).
    end{align*}$$
    Now since $n>3$, this is an immediate contradiction.



    Thanks for all the fast help!






    number-theory polynomials field-theory radicals







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      3



      $begingroup$


      Let $k$ be a field with char$(k)=0$ and suppose for $f,g,hin k[x]$ having gcd$(f,g,h)=1$ and $ninmathbb{N}_{geqslant4}$ it holds that $f^n+g^n=h^2$. I want to show that $f,g,h$ are all constant. Of course, we go the Mason Stothers-way. Quick note: by gcd $1$ and the equation, we have that $f,g,h$ are pairwise co-prime.



      For the sake of contradiction, suppose that $f,g,h$ are not all constant. By the other conditions listed above, we can apply Mason-Stothers like so:
      $$max(deg f^n,deg g^n,deg h^2)leqslantdeg(text{rad}(fgh))-1.$$
      Now let's look at two cases:



      1. $f^n$ has maximal degree (and is non-constant).



      2. $h^2$ has maximal degree.



      A contradiction should appear in both cases. Let's begin with 1.:
      Since $f^n$ has max degree and the polynomials are pairwise co-prime, we can say $$begin{align*}4leqslantdeg(f^n)=ndeg(f)&<deg({text{rad}(fgh)})\
      &= deg(text{rad}(f)text{rad}(g)text{rad}(h))\
      &=deg(text{rad}(f))+deg(text{rad}(g))+deg(text{rad}(h))\
      &leqslant3deg(f).
      end{align*}$$
      Now since $n>3$, this is an immediate contradiction.



      Thanks for all the fast help!






      number-theory polynomials field-theory radicals







      share|cite|improve this question











      $endgroup$




      Let $k$ be a field with char$(k)=0$ and suppose for $f,g,hin k[x]$ having gcd$(f,g,h)=1$ and $ninmathbb{N}_{geqslant4}$ it holds that $f^n+g^n=h^2$. I want to show that $f,g,h$ are all constant. Of course, we go the Mason Stothers-way. Quick note: by gcd $1$ and the equation, we have that $f,g,h$ are pairwise co-prime.



      For the sake of contradiction, suppose that $f,g,h$ are not all constant. By the other conditions listed above, we can apply Mason-Stothers like so:
      $$max(deg f^n,deg g^n,deg h^2)leqslantdeg(text{rad}(fgh))-1.$$
      Now let's look at two cases:



      1. $f^n$ has maximal degree (and is non-constant).



      2. $h^2$ has maximal degree.



      A contradiction should appear in both cases. Let's begin with 1.:
      Since $f^n$ has max degree and the polynomials are pairwise co-prime, we can say $$begin{align*}4leqslantdeg(f^n)=ndeg(f)&<deg({text{rad}(fgh)})\
      &= deg(text{rad}(f)text{rad}(g)text{rad}(h))\
      &=deg(text{rad}(f))+deg(text{rad}(g))+deg(text{rad}(h))\
      &leqslant3deg(f).
      end{align*}$$
      Now since $n>3$, this is an immediate contradiction.



      Thanks for all the fast help!






      number-theory polynomials field-theory radicals




      number-theory polynomials field-theory radicals






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 15 '18 at 22:49







      Algebear

















      asked Dec 15 '18 at 21:56









      AlgebearAlgebear

      704419




      704419






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          We don't know anything useful about $deg(text{rad}(f))$, but we do know that it's at most $deg(f)$.



          Similarly, $deg(text{rad}(g)) le deg(g)$, and since $f^n$ has maximal degree, we know $deg(g) le deg(f)$.



          Finally, $deg(text{rad}(h)) le deg(h)$. We know that $deg(h^2) le deg (f^n)$, so $deg(h) le frac n2 deg(h)$.



          (Be careful: we don't know $deg(h) le deg(f)$ just from $deg(h^2) le deg(f^n)$.)



          In the end, we get $$n deg(f) < deg(f) + deg(f) + frac n2 deg(f)$$ from which we have $n < 4$.



          The second case is similar, except instead we have $deg(f^n) le deg(h^2)$, so $deg (f) le frac 2n deg (h)$. The inequality we get is $2deg(h) < frac2n deg(h) + frac2n deg(h) + deg(h)$ which also implies $n<4$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
            $endgroup$
            – Algebear
            Dec 15 '18 at 22:47





















          1












          $begingroup$

          Since $f^n$ has maximal degree, it follows that $deg(g)leqdeg(f)$. Moreover,
          $$
          2deg(h)leq ndeg(f).
          $$
          Since $operatorname{rad}$ only lowers the degree, the RHS is at most
          $$
          left(2+frac n2right)deg(f),
          $$
          Which is at most $ndeg(f)$ since $ngeq4$.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            We don't know anything useful about $deg(text{rad}(f))$, but we do know that it's at most $deg(f)$.



            Similarly, $deg(text{rad}(g)) le deg(g)$, and since $f^n$ has maximal degree, we know $deg(g) le deg(f)$.



            Finally, $deg(text{rad}(h)) le deg(h)$. We know that $deg(h^2) le deg (f^n)$, so $deg(h) le frac n2 deg(h)$.



            (Be careful: we don't know $deg(h) le deg(f)$ just from $deg(h^2) le deg(f^n)$.)



            In the end, we get $$n deg(f) < deg(f) + deg(f) + frac n2 deg(f)$$ from which we have $n < 4$.



            The second case is similar, except instead we have $deg(f^n) le deg(h^2)$, so $deg (f) le frac 2n deg (h)$. The inequality we get is $2deg(h) < frac2n deg(h) + frac2n deg(h) + deg(h)$ which also implies $n<4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
              $endgroup$
              – Algebear
              Dec 15 '18 at 22:47


















            2












            $begingroup$

            We don't know anything useful about $deg(text{rad}(f))$, but we do know that it's at most $deg(f)$.



            Similarly, $deg(text{rad}(g)) le deg(g)$, and since $f^n$ has maximal degree, we know $deg(g) le deg(f)$.



            Finally, $deg(text{rad}(h)) le deg(h)$. We know that $deg(h^2) le deg (f^n)$, so $deg(h) le frac n2 deg(h)$.



            (Be careful: we don't know $deg(h) le deg(f)$ just from $deg(h^2) le deg(f^n)$.)



            In the end, we get $$n deg(f) < deg(f) + deg(f) + frac n2 deg(f)$$ from which we have $n < 4$.



            The second case is similar, except instead we have $deg(f^n) le deg(h^2)$, so $deg (f) le frac 2n deg (h)$. The inequality we get is $2deg(h) < frac2n deg(h) + frac2n deg(h) + deg(h)$ which also implies $n<4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
              $endgroup$
              – Algebear
              Dec 15 '18 at 22:47
















            2












            2








            2





            $begingroup$

            We don't know anything useful about $deg(text{rad}(f))$, but we do know that it's at most $deg(f)$.



            Similarly, $deg(text{rad}(g)) le deg(g)$, and since $f^n$ has maximal degree, we know $deg(g) le deg(f)$.



            Finally, $deg(text{rad}(h)) le deg(h)$. We know that $deg(h^2) le deg (f^n)$, so $deg(h) le frac n2 deg(h)$.



            (Be careful: we don't know $deg(h) le deg(f)$ just from $deg(h^2) le deg(f^n)$.)



            In the end, we get $$n deg(f) < deg(f) + deg(f) + frac n2 deg(f)$$ from which we have $n < 4$.



            The second case is similar, except instead we have $deg(f^n) le deg(h^2)$, so $deg (f) le frac 2n deg (h)$. The inequality we get is $2deg(h) < frac2n deg(h) + frac2n deg(h) + deg(h)$ which also implies $n<4$.






            share|cite|improve this answer









            $endgroup$



            We don't know anything useful about $deg(text{rad}(f))$, but we do know that it's at most $deg(f)$.



            Similarly, $deg(text{rad}(g)) le deg(g)$, and since $f^n$ has maximal degree, we know $deg(g) le deg(f)$.



            Finally, $deg(text{rad}(h)) le deg(h)$. We know that $deg(h^2) le deg (f^n)$, so $deg(h) le frac n2 deg(h)$.



            (Be careful: we don't know $deg(h) le deg(f)$ just from $deg(h^2) le deg(f^n)$.)



            In the end, we get $$n deg(f) < deg(f) + deg(f) + frac n2 deg(f)$$ from which we have $n < 4$.



            The second case is similar, except instead we have $deg(f^n) le deg(h^2)$, so $deg (f) le frac 2n deg (h)$. The inequality we get is $2deg(h) < frac2n deg(h) + frac2n deg(h) + deg(h)$ which also implies $n<4$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 15 '18 at 22:39









            Misha LavrovMisha Lavrov

            47.4k657107




            47.4k657107












            • $begingroup$
              Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
              $endgroup$
              – Algebear
              Dec 15 '18 at 22:47




















            • $begingroup$
              Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
              $endgroup$
              – Algebear
              Dec 15 '18 at 22:47


















            $begingroup$
            Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
            $endgroup$
            – Algebear
            Dec 15 '18 at 22:47






            $begingroup$
            Ahh, okay. That second case caused the most trouble. Excuse the fact that I added the proof of case 1 after the post. Anyways, your proof is more detailed, thanks.
            $endgroup$
            – Algebear
            Dec 15 '18 at 22:47













            1












            $begingroup$

            Since $f^n$ has maximal degree, it follows that $deg(g)leqdeg(f)$. Moreover,
            $$
            2deg(h)leq ndeg(f).
            $$
            Since $operatorname{rad}$ only lowers the degree, the RHS is at most
            $$
            left(2+frac n2right)deg(f),
            $$
            Which is at most $ndeg(f)$ since $ngeq4$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Since $f^n$ has maximal degree, it follows that $deg(g)leqdeg(f)$. Moreover,
              $$
              2deg(h)leq ndeg(f).
              $$
              Since $operatorname{rad}$ only lowers the degree, the RHS is at most
              $$
              left(2+frac n2right)deg(f),
              $$
              Which is at most $ndeg(f)$ since $ngeq4$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Since $f^n$ has maximal degree, it follows that $deg(g)leqdeg(f)$. Moreover,
                $$
                2deg(h)leq ndeg(f).
                $$
                Since $operatorname{rad}$ only lowers the degree, the RHS is at most
                $$
                left(2+frac n2right)deg(f),
                $$
                Which is at most $ndeg(f)$ since $ngeq4$.






                share|cite|improve this answer









                $endgroup$



                Since $f^n$ has maximal degree, it follows that $deg(g)leqdeg(f)$. Moreover,
                $$
                2deg(h)leq ndeg(f).
                $$
                Since $operatorname{rad}$ only lowers the degree, the RHS is at most
                $$
                left(2+frac n2right)deg(f),
                $$
                Which is at most $ndeg(f)$ since $ngeq4$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 15 '18 at 22:43









                Michael BurrMichael Burr

                26.9k23262




                26.9k23262






























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