Given a polynomial $W(x)$. Prove that there exists an integer $m$ that satisfies $W(m)=W(m+1)=0$. 
Given a polynomial $W(x)=x^2+ax+b$ with integer coefficients that satisfies the condition: for every prime $p$ there exists an integer $k$ that $p$ divides both $W(k)$ and $W(k+1)$. Prove that there exists an integer $m$ that satisfies $W(m)=W(m+1)=0$.

I am asking you to check my reasoning.
We can consider a new polynomial $P(x) = 4W(x)=(2x+a)^2+4b-a^2$. $P(k+1)-P(k)=(2(k+1)+a)^2-(2k+a)^2=4(2k+1+a)$. Consider pairs $(2(k+1)+a,2k+1+a)$ and $(2k+a,2k+1+a)$ but these are pairs of coprime numbers meaning $gcd(P(k),P(k+1))=4$ contradicting the problem's condition. Hence $P(k)=p(k+1)=0$.
 A: 
[I]f both $P(k)$ and $P(k + 1)$ are not a zero polynomial no such prime $r > 2$ divides both $P(k)$ and $P(k + 1)$ simultaneously which means $P(k) = P(k + 1) = 0$.

This statement makes no sense at all. Firstly, if $k$ is a number then $P(k)$ is also a number, not a polynomial. Second, if no prime $r > 2$ divides both $P(k)$ and $P(k + 1)$ simultaneously, then it is impossible that $P(k) = P(k + 1) = 0$, since then all primes would divide both simultaneously.
A better approach is:
Consider $p = 2$. Then we have $W(k) \equiv W(k + 1) \equiv 0 \mod 2$. One of $k$ and $k + 1$ must be even, so this means that $b$ is even. And one of them must be odd, which means that $a$ is odd. Write $a = -(2m + 1)$.
Consider $J(x) = W(x - m)$. Then we see that $J(x) = x^2 - x + c$ for some even $c$ (in fact, it's $c = m^2 + am + b$, which is even). We also see that for all $p$ prime, there exists $k$ such that $J(k) \equiv J(k + 1) \equiv 0 \mod p$. This is because we can take $k'$ such that $W(k') \equiv W(k' + 1) \equiv 0 \mod p$ and set $k = k' + m$.
Now let us suppose that, for some prime $p > 2$, we have $J(k) \equiv J(k + 1) \equiv 0 \mod p$. Then we have $j^2 - j + c \equiv 0 \mod p$ and $(j + 1)^2 - (j + 1) + c \equiv 0 \mod p$. Then in particular, we have $(j + 1)^2 - (j + 1) + c = j^2 + 2j + 1 - j - 1 + c = j^2 - j + c + 2j \equiv 2j \equiv 0 \mod p$. Then $j \equiv 0 \mod p$. Then $c \equiv j^2 + j + c \equiv 0 \mod p$.
So we see that $c$ is divisible by every prime number. Then $c = 0$. Then $J(0) = J(1) = 0$. Then $J(0) = W(m)$, $J(1) = W(m + 1)$. So $W(m) = W(m + 1) = 0$.
A: Such $m$ existy only if $W(x) =x^2-(2m+1)x+m^2+m$

Let $2\mid a$, then for some integer $k$ we have $$2\mid W(k+1)-W(k) = 2k+1+a \implies 2\mid 2k+1$$ A contradiction, so $a$ is odd: $a=-2m-1$ for some integer $m$. Now let $p>2$, then exists $k$ such that $$p\mid 2(k-m)\implies p\mid k-m \implies k\equiv_p m$$
But $$0\equiv_p W(k) = k^2 +k(-2m-1) + b \equiv_p -m^2-m+b $$
So, for every $p$ we have $$b\equiv _pm^2+m$$ But this is only possible if $b=m^2+m$.
