I claim to show that every number appears even times in the sequence $a_1,a_2,\dotsc, a_n$ so $n$ is even. So I erase the repeated numbers Each time I erase exactly 2 equal numbers and I continue until every number appears at most once. Or in the other meaning if $b_1,b_2,\dotsc,b_n$ are distinct numbers that $p(x)=(x+b_1)^{\alpha_1}{(x+b_2)}^{\alpha_2}\dotsb (x+b_n)^{\alpha_n}$ I change the $p(x)$ to $(x+b_1)^{\alpha'_1}{(x+b_2)}^{\alpha'_2}\dotsb (x+b_n)^{\alpha'_n}$ where $\forall i : \alpha'_i \in \{0,1\}$ and $\alpha'_i \overset{2}{\equiv} \alpha_i$ and by the absurd hypothesis the new $n$ is not $0$. Puriya Zarei's love: We have a plank of wood with $n$ integer holes in it, and there is an axis where on every integer point there is a lamp. Every time we can place the wood's holes on integer points (We cannot rotate the wood) and the state of all of the lamps behind the holes will be reversed. If in the first state all of the lamps be off and $n \ge 1$, The lemma states that it's possible to turn on exactly $2$ lamps. Proof: Assume that the length of the wood is $k$ and there is a hole at the left of the wood. If $k=n=1$ it's trivial. So I assume that $k \ge n>1$. Consider this algorithm: At the first move push the wood to an arbitrary place. then every time push the wood's first hole(from the left) on the 2nd lamp from the left. Continue it forever. After applying the wood (for the $i$'th time) the lamps behind the wood have some situations like $S_i$. ($S_i=(s_{i,1},s_{2,i},\dotsc s_{i,k})$) Create a graph It's a specially directed graph that every vertex's outdegree is exactly $1$ and I don't know its English name but in Persian, we call it "گراف تابع". If anybody knows Persian please translate it. with $2^k+1$ vertices that all its vertices have the first element $0$(the first lamp should be off) except the vertex $S_1$. that every vertex of the graph like $v$ is a situation of lamps and its outcoming edge goes to a vertex $u$ when after happening $v$ the next situation can be uniquely defined $u$. The only vertex with in-degree $0$ is $S_1$. If $(S_1, X)$ is an edge of the graph, X will be the only vertex with in-degree $2$. All other vertices have in-degree $1$. Assume that $X$ is in a cycle with size $t$. $S_{t+2}=S_2$ so $S_{t+1}$ and $S_1$ have a difference at just their first bit. So, before applying the $t+1$th move under the wood plank just the first lamp was on. Due to that, it was the 2nd lamp that is on, the lemma has been proved$\blacksquare$ With the lemma the rest is easy. With the lemma, if $n>0$ we can conclude that there exist numbers $b<c$ such that $\forall k: 2 | \Omega(x+b)+\Omega(x+c)$. Plug $x=k(c-b)-b$ in it and get $\forall k>\frac{b}{c-b}: 2 | \Omega(k(c-b))+\Omega((k+1)(c-b)=2\Omega(c-b) + \Omega(k) + \Omega(k+1) \implies \Omega(k+1) \overset{2}{\equiv} \Omega(k)$ and it makes a contradiction with that $\Omega(2k) \overset{2}{\not \equiv} \Omega(k)$. So the new $n$ is 0 and we are done$\dotsc\checkmark\blacksquare$