# The derivatives of a real polynomial function $f(x,y)$ are divisible by a certain polynomial of $x$ and $y$. What can we know about $f\mkern1mu$?

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

• You mean that the derivatives are multiples of $x^2+y^2-1$? Aug 28, 2019 at 20:08
• Yes. The derivatives can be written as $P(x,y)(x^2+y^2-1)$ with $P$ a polynomial Aug 28, 2019 at 20:11
• Should I rewrite the question to make it clearer? Aug 28, 2019 at 20:12
• 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 $x^2+y^2-1$. Aug 28, 2019 at 20:18
• Thanks. I wasn't sure about how to translate it Aug 28, 2019 at 20:19

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.

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

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 $$B$$.

• 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? Aug 29, 2019 at 15:41
• 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 $(x^2+y^2-1)^2$.
– user502266
Aug 29, 2019 at 17:12
• P.S. This division result applies to any polynomial $f(x,y)$ and does not use any of your other assumptions.
– user502266
Aug 29, 2019 at 17:15
• Now I understand, thank you very much! Aug 29, 2019 at 17:58
• You're very welcome.
– user502266
Aug 29, 2019 at 18:12