$f(x^{17}, y^{29})$ is a polynomial combination of $g(x^{17}, y^{29}), h(x^{17}, y^{29}),$ prove $f$ is a polynomial combination of $g, h.$ $f \in S = \mathbb{R}[x,y]$ is a taxicab polynomial if we can write $f(x,y)=g(x^{17},y^{29})$ for some $g$ in $S,$ and we define $T(f)$ as the corresponding $g.$ If $f, g, h$ are taxicab polynomials and $f=rg+sh$ for some $r, s \in S,$ prove $T(f)=uT(g)+vT(h)$ for some $u, v \in S.$
I considered proving $r, s$ are taxicab polynomials, but this is false by considering $r = x, s = 1-x, f = g = h = x^{17}.$ After this idea failed, I'm not sure what to do.
 A: Let $p=17$ and $q=29$.
Let $a=\sum_{i',j'} \alpha_{i',j'} X^{i'}Y^{j'}$ be an arbitrary polynomial in ${\mathbb R}[X,Y]$. For any pair $(i,j)$ with $0\leq i\lt p$ and $0\leq j\lt q$, let
$$C(i,j)=\bigg\lbrace (i',j') \in {\mathbb N}^2 \ \bigg| \ i' \equiv i \ \mod{p}, j' \equiv j \ \mod{q}  \bigg\rbrace \tag{1}$$
Clearly, ${\mathbb N}^2$ is partitioned by the $C(i,j), 0\leq i\lt p, 0\leq j\lt q$. Now let
$$
a_{ij}(X,Y)=\sum_{(i',j')\in C(i,j)} \alpha_{i',j'} X^{\frac{i'-i}{p}}Y^{\frac{j'-j}{q}} \tag{2}
$$
Those $a_{ij}$'s are polynomials in ${\mathbb R}[X,Y]$. By construction, we then have
$$
a=\sum_{i=0}^{p-1}\sum_{j=0}^{q-1} X^iY^j a_{i,j}(X^p,Y^q) \tag{3}
$$
Conversely, if we have a family of polynomials $a_{i,j}$'s satisfying (3), it is clear by identifying coefficients that they must be given by (2). So the decomposition (3) is uniquely defined ;  I call it the taxicab decomposition of $a$.
Now, starting from the identity
$$
f(X^p,Y^q)=r(X,Y)g(X^p,Y^q)+s(X,Y)h(X^p,Y^q) \tag{4}
$$
We now invoke the taxicab decomposition of both $r$ and $s$ :
$$
r(X,Y)=\sum_{i=0}^{p-1}\sum_{j=0}^{q-1} X^iY^j r_{i,j}(X^p,Y^q),\
s(X,Y)=\sum_{i=0}^{p-1}\sum_{j=0}^{q-1} X^iY^j s_{i,j}(X^p,Y^q) \tag{5}
$$
Injecting (5) into (4), we obtain
$$
f(X^p,Y^q)=\sum_{i=0}^{p-1}\sum_{j=0}^{q-1} X^iY^j (r_{i,j}(X^p,Y^q)g(X^p,Y^q)+s_{i,j}(X^p,Y^q)h(X^p,Y^q)) \tag{6}
$$
By uniqueness of the taxicab decomposition, from (6) we deduce that
$$
f(X^p,Y^q)=r_{0,0}(X^p,Y^q)g(X^p,Y^q)+s_{0,0}(X^p,Y^q)h(X^p,Y^q) \tag{7}
$$
Now since the taxicab homomorphism $a(X,Y)\mapsto a(X^p,Y^q)$ is injective, (5) forces us to have
$$
f(X,Y)=r_{0,0}(X,Y)g(X,Y)+s_{0,0}(X,Y)h(X,Y) \tag{8}
$$
which is exactly the kind of thing we're looking for. This finishes the proof.
A: This is tricky. I don't have an answer, but I do have a partial result that constrains some of the coefficients of low degree of our starting $r$ and $s$.
First, I want to reframe the problem. Let $f, g, h$ be polynomials in $\mathbb{R}[x, y]$. I define the anti-taxicab polynomials as $A = \mathbb{R}[x^\frac{1}{17}, y^\frac{1}{29}]$.
The question as stated is equivalent to:
Let $f, g, h$ be polynomials in two variables with real coefficients, suppose that there exist $r, s$ in $A$ such that $f = r*g + s* h$. Prove that there also exists an $r'$ and $s'$ in $\mathbb{R}[x, y]$ satisfying the same equation. I can't prove that, but I can constrain the coefficients of $r$ and $s$.
Consider $x^a$ where $1 \le 17a \le 16$.
$f[x^a]$ must be $0$ because $f$ is a polynomial.
Thus $(r * g + s * h)[x^a]$ must also be zero.
$(r * g + s * h)[x^a] = r[x^a]g[x^0] + s[x^a]h[x^0]$
This means that $r[x^a] = -\frac{s[x^a]h[x^0]}{g[x^0]}$ and $r[x^a] = 0$ when the RHS of the previous expression is undefined.
The same argument works for $y^a$ where $1 \le 29a \le 28$.
