'Completing the square' in higher degrees Let $p_2=ax^2+bxy+cy^2$, $a,b,c \in \mathbb{R}$ with $a \neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain: 
$p_2=a(x^2+\frac{b}{a}xy+\frac{c}{a}y^2)=
a(x^2+2x\frac{b}{2a}y+\frac{b^2}{4a^2}y^2-\frac{b^2}{4a^2}y^2+\frac{c}{a}y^2)=
a((x+\frac{b}{2a}y)^2+(\frac{c}{a}-\frac{b^2}{4a^2})y^2)=
a(x+\frac{b}{2a}y)^2+a\frac{4ac-b^2}{4a^2}y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d \in \mathbb{R}$ with $a \neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+\frac{b}{a}x^2y+\frac{c}{a}xy^2+\frac{d}{a}y^3)=
a(x^3+3x^2\frac{b}{3a}y+3x(\frac{b}{3a}y)^2+(\frac{b}{3a}y)^3
-3x(\frac{b}{3a}y)^2+\frac{c}{a}xy^2-(\frac{b}{3a}y)^3+\frac{d}{a}y^3)=
a(x^3+3x^2\frac{b}{3a}y+3x(\frac{b}{3a}y)^2+(\frac{b}{3a}y)^3)+
a(-3x\frac{b^2}{9a^2}y^2+\frac{c}{a}xy^2-\frac{b^3}{27a^3}y^3+\frac{d}{a}y^3)=
a(x+\frac{b}{3a}y)^3+
a(\frac{c}{a}-3\frac{b^2}{9a^2})xy^2+a(\frac{d}{a}-\frac{b^3}{27a^3})y^3=
a(x+\frac{b}{3a}y)^3+
\frac{9ac-3b^2}{9a}xy^2+\frac{27a^2d-b^3}{27a^2}y^3=
a(x+\frac{b}{3a}y)^3+
\epsilon xy^2+\delta y^3$
where $\epsilon:=\frac{9ac-3b^2}{9a}=\frac{3ac-b^2}{3a}$ and $\delta:=\frac{27a^2d-b^3}{27a^2}$.
Now let $G: (x,y) \mapsto (x-\frac{b}{3a}y,y)$; $G$ is an (affine) automorphism of $\mathbb{R}$. We get that: $G(p_3)=G(a(x+\frac{b}{3a}y)^3+
\epsilon xy^2+\delta y^3)= aG(x+\frac{b}{3a}y)^3+
\epsilon G(xy^2)+\delta G(y^3)=ax^3+
\epsilon (x-\frac{b}{3a}y)y^2+\delta y^3=
ax^3+
\epsilon xy^2+ (\delta-\epsilon\frac{b}{3a})y^3$
Therefore, $\frac{\partial (G(p_3))}{\partial x}=3ax^2+\epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree. 

Is there a similar trick for higher odd degrees $2n+1 \in \{5,7,\ldots\}$? Namely, can one find an affine automorphism $G$ of $\mathbb{R}[x,y]$ such that every monomial in $\frac{\partial G(p_{2n+1})}{\partial x}$ has even $x$ degree and even $y$ degree. ?

The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then 
$\frac{\partial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)}{\partial x}
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
 A: By dehomogenizing the equation, this is equivalent to finding a substitution $x\mapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $\Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $x\mapsto x+c$ is a group action by $\Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_{2n+1}x^{2n+1}+a_{2n-1}x^{2n-1}+\cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
