Find the smallest integer k so that for all quadratic polynomials P with integer coefficients, one of $P(1),\cdots, P(k)$ has a 0 in base 2 (obviously leading zeroes don't count).
Let $\mathcal{P}$ denote the set of all quadratic polynomials with integer coefficients. Say a positive integer k is good if it satisfies that for all $P\in\mathcal{P},$ one of $P(1),\cdots, P(k)$ has a 0 in base 2. Obviously $k=1$ fails because we can choose $P\in\mathcal{P}$ so that $P(1) = 1$ (e.g. $P(x) = x^2$). If $k$ is good, then every $n\ge k$ is obviously good. $k=2$ also fails as we can choose $P(x)=(x-1)(x-2)+1$. That one of $P(1),\cdots, P(k)$ has a zero in base 2 is equivalent to saying that not all of them can be written in the form $\pm (2^s-1)$ for some $s\ge 1$. If $P(i)\neq 0$ has a zero in base 2, it is either a nontrivial power of 2 or $2^j | P(i)$ for some $j\ge 1$ or there must exist $a < b-1$ with $P(i)\equiv x\mod 2^b$ where $x \leq 2^{a+1}-1.$ Indeed it's clear that if the first two cases don't hold then $P(i)$ must be odd. Since it has a 0 in its binary representation, we can pick the smallest index j so that the jth digit in the binary representation is a 1 while the $(j+1)$th is a zero. Then j cannot be the index of the most significant binary digit or leftmost digit of $P(i)$ since $P(i)$ has a zero, and since we know the leftmost digit of $P(i)$ is a 1, there must be some $k\ge j+2$ so that the kth digit of the binary representation of $P(i)$ is a 1. I'm not sure how to proceed from here.
Suppose $P(x) = ax^2 + bx + c$ is a quadratic polynomial with integer coefficients. If $P(i)$ does not have a zero in base 2, it must be of the form $\pm (2^s - 1)$ for some $s\ge 1.$ For $k=3$ we have the counterexample $P(x) = x^2 - x+1$. Note that the latter polynomial is the unique quadratic polynomial satisfying $P(1) = 1, P(2) = 3, P(3) = 7.$ In general if we want to find a quadratic polynomial satisfying $P(i) = a_i$ for $1\leq i\leq 3,$ then the unique polynomial of degree at most 2 satisfying this requirement is $\begin{align}P(x) &= \sum_{i=1}^3 a_i \prod_{j\neq i, 1\leq j\leq 3}\dfrac{(x-j)}{i-j}\\ &= x^2(\dfrac{a_1+a_3}2 - a_2)+x(\dfrac{-5a_1-3a_3}2+4a_2) + (3a_1 - 3a_2+a_3)\\ &= a_1(\dfrac{x^2}2-\dfrac{5}2 x+3) + a_2(-x^2+4x-3) + a_3(\dfrac{x^2}2-\dfrac{3}2x+1)\end{align}\tag{1}$
Let this polynomial be given by $P_{a_1,a_2,a_3}.$ Formula (1) shows that there is a unique quadratic polynomial $P$ with integer coefficients so that $P(i) = a_i$ for $1\leq i\leq 3$ where each $a_i$ is an integer if and only if $a_1\equiv a_3$ mod 2 and $a_2\neq \dfrac{a_1+a_3}2.$ Note that such a quadratic polynomial, if it exists, is unique, so it suffices to show existence. From formula $1$ we get $P(4) = 16((a_1+a_3)/2-a_2)+4((-5a_1-3a_3)/2+4a_2) + (3a_1-3a_2+a_3) = a_1-3a_2 + 3a_3.$ We also get $P(5) = 3a_1-8a_2+6a_3, P(6) = 6a_1-15a_2+10a_3.$ So $k=4$ fails with the polynomial $P_{1,3,3}$ being a counterexample. Note that $k=5$ fails due to this same polynomial. It might be helpful to investigate when $P_{a_1,a_2,a_3}(4)$ is of the form $\pm (2^s-1)$ for some $s\ge 1$ and $a_1,a_2,a_3$ are all of the form $\pm (2^t - 1)$ for some integer $t\ge 1$. Write $a_i = \epsilon_i (2^{s_i} - 1)$ where each $\epsilon_i \in \{1,-1\}$. Then we get $P_{a_1,a_2,a_3}(4) = \epsilon_1(2^{s_1}-1) - 3\epsilon_2 (2^{s_2} - 1) + 3\epsilon_3 (2^{s_3} - 1).$ But I'm not sure how to proceed using this approach.
