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Given $f(x,y)$, a two real variable polynomial such that $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ are divisible by $x^2+y^2-1$. Prove that there exist a polynomial $g(x,y)$ and a constant $c$ such that: $$f(x,y)= g(x,y){(x^2+y^2-1)}^2+c$$

That is the problem I'm currently trying to solve. I first tried to write de derivatives in the form $P(x,y)* ({x^2+y^2-1})$ with $P(x,y)$ being a polynomial. Then tried to write both $P$ and ${x^2+y^2-1}$ as derivatives of a function (one at a time) to perform integrarion by parts, but it led me nowhere.

Any ideas in how to solve it? If it's possible to just give me a hint in how to start and not the whole solution I'd appreciate it very much

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  • $\begingroup$ You mean that the derivatives are multiples of $x^2+y^2-1$? $\endgroup$
    – Aphelli
    Aug 28, 2019 at 20:08
  • $\begingroup$ Yes. The derivatives can be written as $P(x,y)(x^2+y^2-1)$ with $P$ a polynomial $\endgroup$ Aug 28, 2019 at 20:11
  • $\begingroup$ Should I rewrite the question to make it clearer? $\endgroup$ Aug 28, 2019 at 20:12
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    $\begingroup$ Yes; you write "...that $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ divide $x^2+y^2-1$." but in stead you mean "...that $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ are divisible by $x^2+y^2-1$. $\endgroup$ Aug 28, 2019 at 20:18
  • $\begingroup$ Thanks. I wasn't sure about how to translate it $\endgroup$ Aug 28, 2019 at 20:19

2 Answers 2

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We may assume that $f(1,0)=0.$

Note that $$\frac{d}{d\theta}(f(\cos{\theta},\sin{\theta}))=-\sin{\theta}\frac{\partial f}{\partial x}(\cos{\theta},\sin{\theta})+\cos{\theta}\frac{\partial f}{\partial y}(\cos{\theta},\sin{\theta})=0,$$ by the properties of $f,$ so $f$ vanishes on the unit circle. So we can write $f(x,y)=f_1(x,y)(x^2+y^2-1)$.

Now, $$\frac{d}{dr}_{|r=1} (f(r\cos{\theta},r\sin{\theta}))=\cos{\theta}\frac{\partial f}{\partial x}(r\cos{\theta},r\sin{\theta})+\sin{\theta}\frac{\partial f}{\partial y}(r\cos{\theta},r\sin{\theta})=0.$$

On the other hand, $f(r\cos{\theta},r\sin{\theta})=f_1(r\cos{\theta},r\sin{\theta})(r+1)(r-1)$, so the derivative wrt $r$ when $r=1$ is $2f_1(\cos{\theta},\sin{\theta})$, so $f_1$ vanishes also on the unit circle so is divisible by $x^2+y^2-1$, QED.

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  • $\begingroup$ I could not quite follow your argument, how do you know theses first properties about $f$ and its derivative with respect to $\theta$? $\endgroup$ Aug 28, 2019 at 22:36
  • $\begingroup$ I am just using the fact that $\nabla f$ vanishes on the unit circle. $\endgroup$
    – Aphelli
    Aug 29, 2019 at 6:04
  • $\begingroup$ In fact we get $f(x,y)=f_1(x,y)(x^2+y^2-1)+c$ $\endgroup$ Aug 29, 2019 at 6:08
  • $\begingroup$ I wrote that we assumed without loss of generality that $f(1,0)=0$. $\endgroup$
    – Aphelli
    Aug 29, 2019 at 6:17
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First treat $y$ as a constant. Then, by polynomial division, we can write $f(x,y)$ as

$h(x)(x^2+y^2-1)^2+$ A cubic in $x$.

For the derivative of the cubic to be a multiple of $x^2+y^2-1$, the cubic must be $$a(x^3+3(y^2-1)x)+b,$$

where $a$ and $b$ are constants (i.e. functions of $y$ only). Note that the coefficients of $h(x)$ are also functions of $y$.

The only condition which now needs to be satisfied is for the partial derivative of this cubic w.r.t. $y$ to be a multiple of $x^2+y^2-1$.This partial derivative is $$\frac{da}{dy}(x^3+3(y^2-1)x)+6axy+\frac{db}{dy}$$ and, by comparing coefficients of $x$ and $x^3$, we require $$\frac{da}{dy}(y^2-1)=-3a ,\frac{db}{dy}=0.$$ The differential equations have solutions $a=A(y^2-1)^{-\frac{3}{2}}$ and $b=B$, where $A$ and $B$ are (genuine) constants. So the only polynomial solution is obtained when $A=0 $ and therefore the 'cubic in $x$' is just the constant $B$.

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  • $\begingroup$ Why polynomial division gives us a result about $f$ itself and not its derivatives? I mean, can you be more specific about how our assumptions bring this result? $\endgroup$ Aug 29, 2019 at 15:41
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    $\begingroup$ Let $u(x,y)x^4$ be all the terms in $f(x,y)$ which involve $x^4$. Then we can write $u(x,y)x^4=u(x,y)(x^2+y^2-1)^2-u(x,y)(*)$, where * only involves powers of x less than 4. In this way we can reduce the 'remainder' to a cubic in $x$. It's just normal division by the polynomial $(x^2+y^2-1)^2$. $\endgroup$
    – user502266
    Aug 29, 2019 at 17:12
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    $\begingroup$ P.S. This division result applies to any polynomial $f(x,y)$ and does not use any of your other assumptions. $\endgroup$
    – user502266
    Aug 29, 2019 at 17:15
  • $\begingroup$ Now I understand, thank you very much! $\endgroup$ Aug 29, 2019 at 17:58
  • $\begingroup$ You're very welcome. $\endgroup$
    – user502266
    Aug 29, 2019 at 18:12

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