6. A polynomial $P(x)$ is a base-n polynomial if it is of the form $a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{1} x+a_{0}$, where each $a_{i}$ is an integer between 0 and $n-1$ inclusive and $a_{d}>0$. Find the largest positive integer $n$ such that for any real number $c$, there exists at most one base- $n$ polynomial $P(x)$ for which $P(\sqrt{2}+\sqrt{3})=c$.
Answer: 9
Solution: It is equivalent to determine the largest $n$ such that we cannot find two distinct base$n$ polynomials $P_{1}$ and $P_{2}$ such that $P_{1}(\sqrt{2}+\sqrt{3})=P_{2}(\sqrt{2}+\sqrt{3})$. The difference of two base- $n$ polynomials is a polynomial with integer coefficients whose absolute values are less than $n$, and all such polynomials are the difference of two base- $n$ polynomials. We compute the minimal polynomial of $x=\sqrt{2}+\sqrt{3}$ first: since $x^{2}=5+2 \sqrt{6}$, we have $\left(x^{2}-5\right)^{2}=24$ so $x^{4}-10 x^{2}+1=0$. Therefore $\sqrt{2}+\sqrt{3}$ is a root of $\left(x^{2}+1\right)\left(x^{4}-10 x^{2}+1\right)=x^{6}-9 x^{4}-9 x^{2}+1$. The coefficients of this polynomial have magnitude at most 9 , so $n9$. Then
$$
\begin{aligned}
\left|c_{d} y^{d}\right| & \geq y^{d} \\
& >\frac{8}{y-1}\left(y^{d}-1\right) \\
& =8 y^{d-1}+8 y^{d-2}+\cdots+8 y+8 \\
& \geq\left|c_{d-1} y^{d-1}+c_{d-2} y^{d-2}+\cdots+c_{0}\right|
\end{aligned}
$$

Therefore $Q(\sqrt{2}+\sqrt{3})=c_{d} y^{d}+c_{d-1} y^{d-1}+\cdots+c_{0} \neq 0$, so no two distinct base-9 polynomials coincide at $x=\sqrt{2}+\sqrt{3}$.
The same logic applies for the odd polynomial case after dividing out a factor of $\sqrt{2}+\sqrt{3}$, so $n=9$ works.