I am wondering about the existence of projections from a multivariate polynomial $p\in\mathbb{R}[x_1,...x_n]$ to a polynomial $q\in\mathbb{R}[t]$ in the sense that there is a map $\phi:\mathbb{R}\to\mathbb{R}^n$ such that for all $t\in\mathbb{R}$ we have $$q(t)=p(\phi(t)).$$ Basically, this amounts to finding a path $\Phi$ (and parametrizing it) in the $n$-dim. hypersurface $\lbrace y=\phi(x)\rbrace$, which, after a suitable projection $\pi$ to a $2$-dim. subspace gives $\pi(\Phi)=\lbrace \hat{y}=q(t)\rbrace$.
This is equivalent to the following problem: Finding algebraic varieties $X,Y$ with coordinate rings $K[X],K[Y],$ respectively, such that $q\in K[X]$ and $p\in K[Y]$ as well as a regular map $\phi:X\to Y$ such that the map $$\phi^{\ast}:K[Y]\to K[X],\, g\mapsto g\circ \phi$$ satisfies the idenity $$q=\phi^{\ast}(p)=p\circ\phi.$$ In my case, the base field $K=\mathbb{R}$.
But how could I proceed? I looked for example at the two polynomials $$p(x_1,x_2,x_3,x_4)=(x_4 (-x_1 + x_4) + x_2 (-x_1 + x_2) + x_1 (3 x_1 - x_4 - x_2 - x_3) + x_3 (-x_1 + x_3))$$ and $$q(t)=t^2(3-t^2)$$ which arise from some application. I tried to use the term $(3-t^2)$ and compare it with $x_1 (3 x_1 - x_4 - x_2 - x_3)$ but I was not succesful. Attempts include $\phi_1(t)=(1+t^2)$ and $\phi_3(t)=\phi_2(t)=\phi_4(t)=\frac{4t^2}{3}$, which gives $p(\phi(t))=\frac{ (t^2 - 3)^2}{3}\neq q(t).$
The choice of $\mathbb{R}$ as base field is quite arbitrary and could be replaced by any field $\mathbb{F}$, simply the differential equation condition would not longer be relevant. Could someone here give me a pointer how one might check for such a correspondance of polynomials?