Hong Kong competition problems

The only polynomial is $x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1$.

Suppose one of the coefficients of $f_0(x) = f(x)$, say the coefficient of $x^k$, is at least $2$. Define $f_1(x) = f_0(x) - 2x^k + x^{k+1}$. Then $f_1(2) = f_0(2) = 2017$ and $f_1(1) < f_0(1) = 7$. Also, the coefficients of $f_1$ are nonnegative integers. Repeating the same process, since the sequence $f_0(1), f_1(1), \dots$ of nonnegative integers is strictly decreasing, the process must end at some step. Then we obtain a polynomial $g(x)$ with coefficients $0$ or $1$ such that $g(2) = 2017$ and $g(1) \le 7$.

Since the binary representation of $2017$ is $1111100001_{(2)}$, we know that $g$ is uniquely determined, and is given by
$$g(x) = x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1.$$
As $g(1) = 7$, the equality of $g(1) \le 7$ should hold. This means we do not need to carry out any process, and $f$ must be the same as $g$.