Prove that the polynomial is $g(x,y)(x^2 + y^2 -1)^2 + c$












1












$begingroup$


This is from a Brazilian math contest for college students (OBMU):



Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



$$frac{partial f}{partial x}(x,y)$$



$$frac{partial f}{partial y}(x,y)$$



are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



$$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






real-analysis multivariable-calculus contest-math multivariate-polynomial







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    This is from a Brazilian math contest for college students (OBMU):



    Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



    $$frac{partial f}{partial x}(x,y)$$



    $$frac{partial f}{partial y}(x,y)$$



    are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



    $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






    real-analysis multivariable-calculus contest-math multivariate-polynomial







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      This is from a Brazilian math contest for college students (OBMU):



      Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



      $$frac{partial f}{partial x}(x,y)$$



      $$frac{partial f}{partial y}(x,y)$$



      are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



      $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






      real-analysis multivariable-calculus contest-math multivariate-polynomial







      share|cite|improve this question











      $endgroup$




      This is from a Brazilian math contest for college students (OBMU):



      Let $f(x,y)$ be a polynomial in two real variables such that the polynomials



      $$frac{partial f}{partial x}(x,y)$$



      $$frac{partial f}{partial y}(x,y)$$



      are divisible by $x^2+y^2-1$. Prove that there's a polynomial $g(x,y)$ and a constant $c$ such that



      $$f(x,y) = g(x,y)(x^2+y^2 -1)^2 +c$$






      real-analysis multivariable-calculus contest-math multivariate-polynomial




      real-analysis multivariable-calculus contest-math multivariate-polynomial






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 0:01









      Jean Marie

      28.9k41949




      28.9k41949










      asked Nov 19 '18 at 23:34









      Rafael DeigaRafael Deiga

      662311




      662311






















          1 Answer
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          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16











          Your Answer





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          2












          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16
















          2












          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16














          2












          2








          2





          $begingroup$

          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.






          share|cite|improve this answer











          $endgroup$



          Treating $f$ as a polynomial in $(mathbb{R}[y])[x]$, there exist polynomials $p(x,y) in (mathbb{R}[y])[x], q(y), r(y) in mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.



          We have that $frac{partial f}{partial x}(x,y)=(x^2+y^2-1)frac{partial p}{partial x}(x,y)+2xp(x,y)+q(y)$ and $frac{partial f}{partial y}(x,y)=(x^2+y^2-1)frac{partial p}{partial y}(x,y)+2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$. Therefore $2xp(x,y)+q(y)$ and $2yp(x,y)+xq’(y)+r’(y)$ are divisible by $(x^2+y^2-1)$



          As we did earlier, there exist polynomials $s(x,y) in (mathbb{R}[y])[x], t(y),u(y) in mathbb{R}[y]$ such that $p(x,y)=(x^2+y^2-1)s(x,y)+xt(y)+u(y)$.



          We have that
          begin{align}
          2xp(x,y)+q(y)&=2x(x^2+y^2-1)s(x,y)+2x^2t(y)+2xu(y)+q(y) \
          &=(x^2+y^2-1)(2xs(x,y)+2t(y))+x(2u(y))+(q(y)-2(y^2-1)t(y))\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $2u(y)=0,q(y)-2(y^2-1)t(y)=0$. Have $q’(y)=4yt(y)+2(y^2-1)t’(y)$



          Next we have
          begin{align}
          2yp(x,y)+xq’(y)+r’(y)&=2y(x^2+y^2-1)s(x,y)+2xyt(y)+2yu(y)+xq’(y)+r’(y)\
          &=2y(x^2+y^2-1)s(x,y)+x(2yt(y)+4yt(y)+2(y^2-1)t’(y))+r’(y)\
          end{align}
          is divisible by $x^2+y^2-1$. Thus $6yt(y)+2(y^2-1)t’(y)=0, r’(y)=0$. Thus $r(y)=c$ for some constant $c$.



          We shall show $t(y)=0$. Assume on the contrary $t$ is not identically $0$. Let $t$ have degree $n$ with nonzero leading coefficient $a$. Comparing the leading coefficient in $6yt(y)+2(y^2-1)t’(y)=0$, we get $6a=2an$, so $n=3$. Note $t$ is divisible by $y^2-1$, so $t(y)=a(y^2-1)(y+b)$, some $b in mathbb{R}$. Substituting $y=0$ in $6yt(y)+2(y^2-1)t’(y)=0$ gives $t’(0)=0$. However $t’(0)=-a$ is nonzero, a contradiction. Thus $t(y)=0$, and so $q(y)=2(y^2-1)t(y)=0$.



          Thus
          begin{align}
          f(x,y)&=(x^2+y^2-1)p(x,y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+x(x^2+y^2-1)t(x,y)+(x^2+y^2-1)u(y)+xq(y)+r(y)\
          &=(x^2+y^2-1)^2s(x,y)+c\
          end{align}
          and we are done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 2:58

























          answered Dec 30 '18 at 2:53









          user630376user630376

          813




          813












          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16


















          • $begingroup$
            When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
            $endgroup$
            – Rafael Deiga
            Dec 30 '18 at 11:16
















          $begingroup$
          When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
          $endgroup$
          – Rafael Deiga
          Dec 30 '18 at 11:16




          $begingroup$
          When you gonna prove that $t(y) =0$, actually we have $6a + 2an =0$, which implies much more easier that $a =0$ xD. The rest is correct i think. Could you fix that?
          $endgroup$
          – Rafael Deiga
          Dec 30 '18 at 11:16


















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