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hint only

$p(x,y)\in\mathbb{R}[x,y]$ and $\forall$ $x,y\in\mathbb{R}$ we have $p(x-1,y-2x+1)=p(x,y)$. prove $\exists$ $f\in\mathbb{R}[x]$ so that $p(x,y)=f(y-x^2)$

my attempt :

if $p(x,y)=f(y-x^2)$ then $p(x,x^2)=f(0)=c$ (this should be correct)

according to the assumption of the problem $p(x,x^2)=p(x-1,(x-1)^2)$ so if $g(x)=p(x,x^2)$ then $g(x)=g(x-1)$ therefor it is clear that $p(x,x^2)=g(x)=c$ (c is constant)

now $$p(x,y)=(y-x^2)p_1(x,y)+r(x)$$ then $$p(x,x^2)=r(x)$$ but we know that $p(x,x^2)=c$ so $r(x)=c$

now $$p(x,y)=p(x-1,y-2x+1)$$ $$(y-x^2)p_1(x,y)+c=(y-2x+1,(x-1)^2)p_1(x-1,y-2x+1)+c$$ $$(y-x^2)\big(p_1(x-1,y-2x+1)-p_1(x,y)\big)=0$$ if $x^2\not=y$ then $p_1(x-1,y-2x+1)=p_1(x,y)$

if $x^2=y$ .....?

farnood gholampoor
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  • It suffices to show that $p(x,y)$ depends only on $y-x^2,$ i.e. if $y-x^2=v-u^2$ then $p(x,y)=p(u,v).$ – Ryszard Szwarc Aug 05 '23 at 07:12
  • @RyszardSzwarc That would be enough to show there is a function $f$. You need some more argumentation to show that $f$ is actually a polynomial. – Arthur Aug 05 '23 at 07:22
  • EDIT: deleted, is probably more explicit of a hint than appropriate – user757173 Aug 05 '23 at 07:25
  • @Arthur Sorry, I was not familiar with the notation $\mathbb{R}[x,y].$ Clearly the conclusion does not hold for any function. So it suffices to show that $p(x,y)=(y-x^2)q(x)+c.$ Then $q$ is a polynomial. – Ryszard Szwarc Aug 05 '23 at 07:25
  • @RyszardSzwarc $P(x,y)=(y-x^2)p_1(x,y)+c_0=(y-x^2)\big((y-x^2)p_2(x,y)+c_1\big)+c_0=\dots=c_n(y-x^2)^n+c_{n-1}(y-x^2)^{n-1} + \dots + c_1(y-x^2)+c_0=f(x-y^2)$

    and we're done for $x^2\not=y$

     – farnood gholampoor Aug 05 '23 at 08:17
  • @farnoodgholampoor Yes, that what I meant. See the answer below. – Ryszard Szwarc Aug 05 '23 at 08:47
  • For what it's worth, the solution I was thinking (which I thought was what Ryszard was getting at) was: $p(x, x^2+l) = p(x-1, (x-1)^2+l)$ and similarly as in the asker's approach, it follows that $p(x, x^2+l) = f(l)$ for some function $f$. Then a substitution gives $p(x, y) = f(y - x^2)$. And plugging in $x=0$ implies that $f$ is a polynomial. – user757173 Aug 05 '23 at 09:20

2 Answers2

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Let $T(x,y)=(x-1,y-2x+1).$ Then $p\circ T=p.$ For $g(x,y)=y-x^2$ we have $g\circ T=g.$ Following OP let $$p(x,y)=g(x,y)p_1(x,y)+r(x)\qquad (1)$$ Then $$p(x,y)=(p\circ T)(x,y)=g(x,y)(p_1\circ T)(x,y)+r(x-1) \quad (2)$$ Subtracting $(1)$ and $(2)$ implies that $r(x)-r(x-1)$ is divisible by $g(x,y)=y-x^2,$ i.e. $r$ is a constant. We thus have $$p(x,y)=g(x,y)p_1(x,y)+r$$ and $p_1(x,y)={p(x,y)-r\over g(x,y)}.$ Therefore $p_1\circ T=p_1.$ We can repeat the same procedure to $p_1(x,y)$ and so on. More formally we can apply the induction argument with respect to the degree of $y.$ Eventually we obtain that $p(x,y)=a_n(y-x^2)^n+\ldots +a_1(y-x^2)+a_0.$

Ryszard Szwarc
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Since $p_1$ satisfies the same condition as $p$ for all $y\ne x^2$ then it satisfies this condition for all $x,y$ and thus $p_1(x,y) = f_1(y-x^2)$ for some polynomial $f_1$ by induction on the degree. So $p(x,y)= f(y-x^2)$ for some polynomial $f$.

nonuser
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