Victorious sets!

If $V$ is a victorious set containing positive integers only, and if $P(x)$ is a polynomial with the defining property for $V$ , then we can find a polynomial $A(x)$ with integer coefficients such that $P(x) - 2x \; = \; A(x) \prod_{v \in V}(x-v)$ and we must have (putting $x=0$ ) $330 \; = \; A(0) \prod_{v \in V}(-v)$ which means that $\prod_{v \in V}v$ must divide $330$ . Thus $V$ must be a subset of the set $\mathcal{V} \; = \; \{ 1,2,3,5,6,10,11,15,22,30,33,50,66,110,165,330\}$ of positive divisors of $330$ , and the product of all the elements of $V$ must also divide $330$ . Without listing them all, there are $103$ such sets. Thus there are $103$ victorious sets consisting of positive integers only, and (by symmetry) there are $103$ victorious sets consisting of negative integers only.

Suppose that $V$ is a victorious set that contains both positive and negative integers, and let $P(x)$ be its associated polynomial. Let $V_+$ and $V_-$ be the positive and negative elements of $V$ respectively. It follows that we can find polynomials $A(x),B(x)$ such that $P(x) - 2x \; = \; A(x) \prod_{v \in V_+} (x -v) \hspace{2cm} P(x) + 2x \; = \; B(x) \prod_{v \in V_-} (x -v)$ and hence $2P(x) \; = \; A(x) \prod_{v \in V_+} (x -v) + B(x) \prod_{v \in V_-} (x -v)$ If $w \in V_-$ then $-4w \; = \; A(w) \prod_{v \in V_+}(w - v)$ so that $\prod_{v \in V_+}(w-v)$ divides $4w$ for any $w \in V_-$ . Similarly, $\prod_{v \in V_-}(w-v)$ divides $4w$ for any $w \in V_+$ . Checking the possibilities, it follows that there are only three types of victorious set $V$ that contain both positive and negative integers: $\{v,-v\}$ , $\{v,-3v\}$ and $\{3v,-v\}$ .

• If $\{v,-v\}$ is a victorious set, then we have $P(x) - 2v \; = \; C(x)(x - v^2)$ for some polynomial $C(x)$ , which implies that $v^2$ must divide $330 - 2v$ . This means that $v$ can be one of $1,3,165$ only. There are $3$ victorious sets like this.
• If $\{v,-3v\}$ is a victorious set, then we have $P(x) = 3v - x + C(x)(x-v)(x+3v)$ for some polynomial $C(x)$ , and hence $v^2$ must divide $110 - v$ . This means that $v$ must be one of $1,2,10,110$ . There are $4$ victorious sets like this.
• By symmetry, there are also $4$ victorious sets of the form $\{3v,-v\}$ .

Thus there are $103 + 103 + 3 + 4 + 4 = \boxed{217}$ different victorious sets.