A polynomial equation game Alice and Bob are sitting in a warm, cozy room. As cozy as the room is, there's not much to do, and they soon find themselves rather bored. 
As you might expect from people as uncreatively named as Alice and Bob, Alice and Bob decide to play a game. They start with a monic polynomial equation $x^n + a_1 x^{n-1} + \cdots + a_n$, where the coefficients $a_i$ are to be determined.
Each turn, Alice chooses an integer, and Bob chooses which coefficient that integer is to be. After choosing and placing all $n$ coefficients, if the resulting polynomial has $n$ distinct integer solutions, Alice wins. Otherwise, Bob wins.
For which $n$ does Alice have a winning strategy? For $n = 1, 2$ it is fairly easy to see she does. What happens for larger $n$?
 A: This is about how I found Alice's strategy for $n=3$, which would be too long for a comment.
Alice's first choice of $0$ was arbitrary. It was chosen to reduce the amount of the calculations. Let polynomial's root to be $p, q, r$.
If bob chooses $a_3=0$, then Alice can adapt the strategy for $n=2$.
If bob chooses $a_2=0$, we know that $pq+qr+rp=0$, and we need to find possible values of $pqr$ and $p+q+r$ and find an integer which can be both. To do this, we need to find some integer solutions of $pq+qr+rp=0$. By trial and error, I found $(km(m+1), k(m+1), -km)$ can be solutions for all integer $k, m$. Then, we get$$pqr=-k_1^3m_1^2(m_1+1)^2, p+q+r=k_2(m_2^2+m_2+1)$$ and it is unlikely that $m^2+m+1$ have many prime factors, so I just chose $m_2=2$ ($1$ or $0$ will make solutions not unique). Then one of $k_1$, $m_1$, $m_1+1$ must be multiple of $7$, so I chose $k_1=1$, $m_1=6$ then $k_2=-252$. Now we have $-k_1^3m_1^2(m_1+1)^2=k_2(m_2^2+m_2+1)=-1764$ and Alice can choose $-1764$ next turn. Alice's choice of final turn is trivial because we already computed the integer solution candidates of polynomial.
If bob chooses $a_1=0$, we know that $p+q+r=0$ or $r=-p-q$. We need to find values for $pq+qr+rp=-p^2-pq-q^2$ and $pqr=-pq(p+q)$, or solve the equation$$-p_1^2-p_1q_1-q_1^2=-p_2q_2(p_2+q_2)$$For convenience, I chose $q_1=0$ and tried to find how to make RHS perfect square. Let $q_2=kp_2$, then we get $p_1^2=k(k+1)p_2^3$. Now I chose the smallest numbers come in mind, that is, $k=2$, $p_2=6$ and $p_1=36$. Then we get $-p_1^2-p_1q_1-q_1^2=-p_2q_2(p_2+q_2)=-1296$.
