If a univariate polynomial is greater than another, is their difference a square?












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Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?






polynomials







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  • $begingroup$
    There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
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    – An_876_Joke
    Jan 4 at 19:18








  • 1




    $begingroup$
    You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
    $endgroup$
    – Martin R
    Jan 4 at 19:21


















0












$begingroup$


Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?






polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
    $endgroup$
    – An_876_Joke
    Jan 4 at 19:18








  • 1




    $begingroup$
    You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
    $endgroup$
    – Martin R
    Jan 4 at 19:21
















0












0








0





$begingroup$


Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?






polynomials







share|cite|improve this question









$endgroup$




Let $f(x)$ and $g(x)$ be polynomials, and suppose that for all $x$, $f(x) ge g(x)$. Does this imply that there exists a polynomial $h(x)$ such that $f(x) - g(x) = h(x)^2$? If so, is this easy to prove, and if not, is there a simple counterexample?






polynomials




polynomials






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asked Jan 4 at 19:13









user54038user54038

1508




1508












  • $begingroup$
    There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
    $endgroup$
    – An_876_Joke
    Jan 4 at 19:18








  • 1




    $begingroup$
    You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
    $endgroup$
    – Martin R
    Jan 4 at 19:21




















  • $begingroup$
    There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
    $endgroup$
    – An_876_Joke
    Jan 4 at 19:18








  • 1




    $begingroup$
    You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
    $endgroup$
    – Martin R
    Jan 4 at 19:21


















$begingroup$
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
$endgroup$
– An_876_Joke
Jan 4 at 19:18






$begingroup$
There are two polynomials $h_1$ and $h_2$ sucht that $f(x) - g(x)= h_1(x)^2 + h_2(x)^2$ since $f(x) - g(x)$ is a non-negative polynomial.
$endgroup$
– An_876_Joke
Jan 4 at 19:18






1




1




$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21






$begingroup$
You can represent the (non-negative) difference as the sum of two squares, see for example math.stackexchange.com/q/1012733/42969.
$endgroup$
– Martin R
Jan 4 at 19:21












3 Answers
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$begingroup$

Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.






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    $begingroup$

    There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$



      so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,



      if that's what you mean..






      share|cite|improve this answer









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        3 Answers
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        3 Answers
        3






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        active

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        $begingroup$

        Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.






            share|cite|improve this answer











            $endgroup$



            Take $f(x)=x^2+2$ and $g(x)=2x$. Then, for each $xinmathbb R$,$$f(x)-g(x)=x^2-2x+2=(x-1)^2+1>0.$$However, there is no polynomial $h(x)$ such that $h^2(x)=x^2-2x+2$ because the degree of $h(x)$ could only be $1$ and therefore it would have to have a real root. But $x^2-2x+2$ has none.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 4 at 19:41

























            answered Jan 4 at 19:19









            José Carlos SantosJosé Carlos Santos

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            165k22132235























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                $begingroup$

                There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.






                    share|cite|improve this answer









                    $endgroup$



                    There should be a very simple counterexample; $f(x)=x^4+x^2+1$ and $g(x)=x^4$. Then $f(x)-g(x)=x^2+1$, which is not a square.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 19:19









                    ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro

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                        1












                        $begingroup$

                        I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$



                        so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,



                        if that's what you mean..






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$



                          so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,



                          if that's what you mean..






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$



                            so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,



                            if that's what you mean..






                            share|cite|improve this answer









                            $endgroup$



                            I don't think so, for example if you take $$f(x)=x^2>g(x)=x-1$$



                            so you can see: $$f(x)-g(x)=x^2-x+1$$ and you cant find $$h(x)^2 $$which equal to that,



                            if that's what you mean..







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 4 at 19:21









                            Yanir ElmYanir Elm

                            693




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