HMMT February, 2020

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 $x^{4} - 10x^{2} + 1 = 0$. Therefore $\sqrt{2} + \sqrt{3}$ is a root of $\left(x^{2} + 1\right)\left(x^{4} - 10x^{2} + 1\right) = x^{6} - 9x^{4} - 9x^{2} + 1$. The coefficients of this polynomial have magnitude at most $9$, so $n < 10$.

In the other direction, observe that $(\sqrt{2} + \sqrt{3})^{k}$ is of the form $a + b\sqrt{6}$ for integers $a$ and $b$ if $k$ is even, and $a\sqrt{2} + b\sqrt{3}$ if $k$ is odd. As no integer linear combination of the first expression can equal the second, we can treat these cases separately. Suppose $Q(x) = c_{d} x^{2d} + c_{d-1} x^{2d-2} + \cdots + c_{0}$ is an even polynomial with $|c_{i}| < 9$ for all $i$ and $c_{d} \neq 0$. Let $y = (\sqrt{2} + \sqrt{3})^{2} = 5 + 2\sqrt{6}$ and observe that $y > 9$. Then
$$\begin{aligned} |c_{d} y^{d}| & \geq y^{d} \\ & > \frac{8}{y-1}(y^{d} - 1) \\ & = 8y^{d-1} + 8y^{d-2} + \cdots + 8y + 8 \\ & \geq |c_{d-1} y^{d-1} + c_{d-2} y^{d-2} + \cdots + c_{0}| \end{aligned}$$
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.