I recently saw a problem that asked you to define an integer-coefficient polynomial that returned an integer for all inputs of the form $1-\sqrt[3]{x}$, where $x$ is an integer
The way I solved it is as follows:
The difference of cubes formula states that $a^3-b^3 = (a-b)*(a^2+ab+b^2)$
In the context of the problem, $a = 1$, and $b = \sqrt[3]{x}$. If we assign $u=1-\sqrt[3]{x}$, we get that:
$1-x=u*(1+\sqrt[3]{x}+\sqrt[3]{x^2})$
We now think of $(1+\sqrt[3]{x}+\sqrt[3]{x^2})$ as some function of $u$, where $1-\sqrt[3]{x}$ has already been substituted in (I guess that would make it a composite function). We know we need a term where $\sqrt[3]{x}$ is squared, so we can first try $(1-\sqrt[3]{x})^2$, which gives us:
$1-2\sqrt[3]{x}+\sqrt[3]{x^2}$
We can now subtract this from our target expression $1+\sqrt[3]{x}+\sqrt[3]{x^2}$ to see how far off we are:
$1+\sqrt[3]{x}+\sqrt[3]{x^2}-(1-2\sqrt[3]{x}+\sqrt[3]{x^2})=3\sqrt[3]{x}$
We can now look for some function of $(1-\sqrt[3]{x})$ that gives us $3\sqrt[3]{x}$, which ends up being:
$-3(1-\sqrt[3]{x})+3$
Substituting $u$ back in, we get that
$1-x=u(u^2-3u+3)$, which gives us our final polynomial of:
$f(u)=u^3-3u^2+3u$ (I'm using $u$ to avoid confusion because I used $x$ in a different context earlier)
After solving this, I tried to solve the similar problem of defining an integer-coefficient polynomial that for all positive integers $x$, $P(\sqrt{x}-\sqrt[3]{x})$ returns an integer. There is no formula I know of that relates $\sqrt{x}$ and $\sqrt[3]{x}$ in a way analogous to the difference of cubes formula, so I wasn't able to approach it from that angle. I tried writing out what $(\sqrt{x}-\sqrt[3]{x})^p$ is for small values of $p$, but that got me nowhere, because there are so many possibilities for the powers of $x$ (All that is required for it to show up is for the power to be of the form $\frac a2 + \frac b3$). I don't have enough math/number theory under my belt to go about solving if this is even possible or not, so I don't know if I'm just wasting time. Any help/solutions would be greatly appreciated!
There is no non-constant polynomial $P(u)$ (note $P(u) = c$, for any integer $c$, always returns an integer for any input), with integer coefficients, where for all positive integers $x$ we get $P(\sqrt{x} - \sqrt[3]{x})$ returning an integer. To see why, first consider the basic underlying reason you were successful with $1 - \sqrt[3]{x}$. With your $f(u)$, the highest order term is $u^3$, which with $u = 1 - \sqrt[3]{x}$ gives $(1 - \sqrt[3]{x})^3 = -x + 3(\sqrt[3]{x})^2 - 3\sqrt[3]{x} + 1$. Since the $x$ and $1$ terms are already integers, for the result to always be an integer, you just need to eliminate the $3(\sqrt[3]{x})^2$ and $- 3\sqrt[3]{x}$ terms, which can be done by adding an appropriate linear combination of $u^2$ and $u$, with it being $-3u^2 + 3u$ in this case.
Assume there's a non-constant integer coefficient polynomial $P(u)$ where, with $u = \sqrt{x} - \sqrt[3]{x} = x^{\frac{3}{6}} - x^{\frac{2}{6}}$, we get $P(u)$ always being an integer for all positive integers $x$. Note that, unlike for $u = 1 - \sqrt[3]{x}$, we can't eliminate the highest non-integer power terms of $x$ in $P(x^{\frac{3}{6}} - x^{\frac{2}{6}})$ by adding a linear combination of smaller degree terms. With the polynomial degree being $d \ge 2$ (note using $x = p^2$ or $x = p^3$, with $p$ prime, proves a linear, non-constant polynomial doesn't work), then expanding $(x^{\frac{3}{6}} - x^{\frac{2}{6}})^{d}$ using the binomial theorem gives terms of $x^{\frac{m}{6}} \; \forall \; 2d \le m \le 3d$.
Since the next highest power term, i.e., $u^{d-1}$, when expanded only has terms in powers of $x$ up to $x^{\frac{3(d - 1)}{6}}$, the $3$ largest power terms of $x^{\frac{3d - 2}{6}}$, $x^{\frac{3d - 1}{6}}$ and $x^{\frac{3d}{6}}$ cannot be eliminated by using a linear combination of lower order powers of $u$. Note at most one of these $3$ terms can be an integral power of $x$, so at least $2$ must be a non-integral power. Also, at least one of $3d - 2 \bmod 6$, $3d - 1 \bmod 6$ and $3d \bmod 6$ must be an odd integer, i.e., one of $1$, $3$ or $5$.
Collect all of the terms with the same fractional power and factor out the integral powers of $x$ to form integer coefficient polynomials in $x$, with $P_i(x)$ for the fractional power of $\frac{i}{6}$. This gives
$$Q_1(x) = P_0(x) + P_1(x)x^{\frac{1}{6}} + P_2(x)x^{\frac{2}{6}} + P_3(x)x^{\frac{3}{6}} + P_4(x)x^{\frac{4}{6}} + P_5(x)x^{\frac{5}{6}} \tag{1}\label{eq1A}$$
Since $P_0(x)$ is always an integer for all positive integers $x$, for $Q_1(x)$ to always be an integer means the remaining terms must add to an integer, so this gives a function $N_1(x) = Q_1(x) - P_0(x)$ which is always an integer, i.e.,
$$N_1(x) = P_1(x)x^{\frac{1}{6}} + P_2(x)x^{\frac{2}{6}} + P_3(x)x^{\frac{3}{6}} + P_4(x)x^{\frac{4}{6}} + P_5(x)x^{\frac{5}{6}} \tag{2}\label{eq2A}$$
Let $x$ be a perfect cube, but not a perfect square, i.e., it's not a perfect sixth power (e.g., $x = p^3$ with $p$ prime). This means $x^{\frac{2}{6}}$ and $x^{\frac{4}{6}}$ are integers, but $x^{\frac{1}{6}}$, $x^{\frac{3}{6}}$ and $x^{\frac{5}{6}}$ are irrational. Also, as indicated earlier about the highest powers of $x$, at least one of $P_1(x)$, $P_3(x)$ and $P_5(x)$ is non-zero, say it's $P_1(x)$. Since the Fundamental theorem of algebra states $P_1(x)$ has a finite number of roots, also have $x$ be one of the infinite number of perfect cubes (but not a perfect square) which are not one of those roots. This gives that $N_2(x) = N_1(x) - P_2(x)x^{\frac{2}{6}} - P_4(x)x^{\frac{4}{6}}$ is an integer, i.e.,
$$N_2(x) = P_1(x)x^{\frac{1}{6}} + P_3(x)x^{\frac{3}{6}} + P_5(x)x^{\frac{5}{6}} \tag{3}\label{eq3A}$$
Move the $P_1(x)x^{\frac{1}{6}}$ term to the left side and square both sides to get
$$\begin{equation}\begin{aligned} (N_2(x) - P_1(x)x^{\frac{1}{6}})^2 & = (P_3(x)x^{\frac{3}{6}} + P_5(x)x^{\frac{5}{6}})^2 \\ N_2^2(x) - 2N_2(x)P_1(x)x^{\frac{1}{6}} + P_1^2(x)x^{\frac{2}{6}} & = P_3^2(x)x + 2P_3(x)P_5(x)x^{\frac{8}{6}} + P_5^2(x)x^{\frac{10}{6}} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
On the right side, all $3$ terms are integers, so their sum is an integer. On the left side, the first & third terms are also integers, so the middle term, i.e., $2N_2(x)P_1(x)x^{\frac{1}{6}}$, must be an integer. However, $x^{\frac{1}{6}}$ is irrational while $2N_2(x)P_1(x)$ is an integer, so it must be $0$. Since $P_1(x) \neq 0$ (as $x$ is not a root), this means $N_2(x) = 0$. You also get $N_2(x) = 0$ if you used $P_3(x)$ or $P_5(x)$ instead of $P_1(x)$.
Use $y = x^{\frac{1}{6}}$ on the right side of \eqref{eq3A} to get
$$Q_2(y) = P_1(y^6)y + P_3(y^6)y^3 + P_5(y^6)y^5 \tag{5}\label{eq5A}$$
The $3$ terms on the right have powers of $y$ which are congruent modulo $6$ to $1$, $3$ and $5$, respectively, so the powers of $y$ in each term are different from any in the other $2$ terms. Since at least one of $P_1(y^6)$, $P_3(y^6)$ and $P_5(y^6)$ is a non-zero polynomial, $Q_2(y)$ must also be a non-zero polynomial. However, it was shown previously that $N_2(x) = Q_2(y)$ is $0$ for an infinite number of values, which contradicts the fundamental theorem of algebra's statement there can only be a finite number of roots. This means the original assumption, i.e., an integer coefficient non-constant $P(u)$ polynomial exists which returns an integer for all $u = \sqrt{x} - \sqrt[3]{x}$ where $x$ is a positive integer, must be false.