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?






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  • 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.
    – An_876_Joke
    Jan 4 at 19:18








  • 1




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


















0














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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  • 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.
    – An_876_Joke
    Jan 4 at 19:18








  • 1




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
















0












0








0







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













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

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  • 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.
    – An_876_Joke
    Jan 4 at 19:18








  • 1




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




















  • 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.
    – An_876_Joke
    Jan 4 at 19:18








  • 1




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


















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.
– An_876_Joke
Jan 4 at 19:18






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.
– An_876_Joke
Jan 4 at 19:18






1




1




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






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












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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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    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.






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      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..






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        3 Answers
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        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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          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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            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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            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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            edited Jan 4 at 19:41

























            answered Jan 4 at 19:19









            José Carlos SantosJosé Carlos Santos

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                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.






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                  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.






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                    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.






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                    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.







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                    answered Jan 4 at 19:19









                    ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro

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                        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








                        New contributor




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                          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








                          New contributor




                          Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





















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                            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








                            New contributor




                            Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            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








                            New contributor




                            Yanir Elm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






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                            answered Jan 4 at 19:21









                            Yanir ElmYanir Elm

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