prove each $a_i$ and $b_j$ equals 0 or 1 
Let $r,s\ge 1$ be integers and $a_0,a_1,\cdots, a_{r-1}, b_0,b_!,\cdots, b_{s-1}$ be real nonnegative numbers such that $(a_0 + a_1x+a_2x^2+\cdots + a_{r-1}x^{r-1} +x^r) (b_0 + b_1x+\cdots + x^s)= 1+x+x^2 + \cdots + x^{r+s}$. Prove that each $a_i$ and each $b_j$ equals 0 or 1.

Note that $a_0b_0 = 1$ and $a_0b_1+a_1b_0 = 1$ implies that $a_0,b_0\leq 1\Rightarrow a_0 = b_0 = 1.$ Clearly $r=s$ is impossible as otherwise the coefficient of $x^r$ is at least $a_0+b_0 = 2$. WLOG (the other case is similar), assume $r < s$. Also the coefficient of $x^k$ in the final product is equal to $\sum_{i=0}^k a_i b_{k-i},$ where $a_i = 0$ for $i > r, a_r=1, b_i = 0$ for $i > s, b_s=1$. All the $a_i$'s and $b_j$'s have magnitude at most 1 but even given $a_0=b_0=1$, I'm not sure how to prove $a_1, b_1$ are both either 0 or 1. It could be useful to use some sort of inequalities (e.g. if $a_i\leq M_i$ for all i and $\sum_{i=1}^{r-1} a_i = \sum_{i=1}^{r-1} M_i,$ then $a_i = M_i$ for all i).
 A: It does not seem that the official solution is wrong. I think it is just incomplete. Anyway, I am going to post a full answer. I hope it is helpful.
First of all, notice that each coefficient should be equal to or less than $1$. The reason is easy. If, for example $a_i \gt 1$, then the coefficient of $x^{i+s}$ is greater than $1$, which is a contradiction. Hence $a_0b_0=1$ implies $a_0=b_0=1.$ By considering the coefficient of $x$, we conclude $a_0b_1+a_1b_0=1 \implies a_1+b_1=1.$ In a similar way, by considering the coefficient of $x^{r+s-1}$, we conclude $ a_{r-1}+b_{s-1}=1.$
Now, by computing the coefficients of $x^s$ and $x^r$, we get:
$$a_0+a_1b_{s-1}+a_2 b_{s-2}+\ ... + =1,$$
and;
$$b_0+b_1a_{r-1}+b_2 a_{r-2}+\ ... + =1.$$
Since $a_0=b_0=1$, the relations above imply that:
$$a_ib_j=0 \    \  if\       \ j+i=s \\ a_ib_j=0\   \   if\   \ i+j=r.$$
Again, since $a_0=b_0=1$ and $a_1+b_1=1$, by the two relations above, if $a_1, b_1 \lt1$, then we should have $a_{r-1}=b_{s-1}=0$, which contradicts $a_{r-1}+b_{s-1}=1$. Therefore $a_1+b_1=1$ and $a_1b_1=0$, which means one of $a_1, b_1$ is one and the other one is zero. Moreover, we obtain that $a_1=a_{r-1}$ and $b_1=b_{s-1}$.
At this stage, let's use strong induction. Assume for every $i\lt k\le \min (r, s)-1$ we have: $a_i=a_{r-i}$ and $b_i=b_{s-i}$  and $a_i, b_i \in \{0,1\}.$ By computing the coefficients of $x^k$ and $x^{r+s-k}$ we get:
$$1=a_0b_k+a_1b_{k-1}+\ ... \ + a_k b_0 ,$$
and:
$$1=b_{s-k}+ b_{s-k+1}a_{r-1} + \ ... \ +a_{r-k};$$
hence $a_k+b_k \in \{0,1\}$ because $a_ib_{k-i}$, by our assumption, is either $0$ or $1$. Similarly, we will have: $a_{r-k}+b_{s-k}=a_k+b_k \in \{0,1\}$, which can be concluded from $a_i=a_{r-i}$ and $b_i=b_{s-i}$. This, together with:
$$a_ib_j=0 \    \  if\       \ j+i=s \\ a_ib_j=0\   \   if\   \ i+j=r;$$
shows that $a_k, b_k \in \{0,1\}$, and $a_k=a_{r-k}$ and $b_k=b_{s-k}$, which completes our claimed induction.
Of course, this argument is not complete (unless $r=s$), and we just showed the coefficients of terms  with powers less than $\min (s,r)$ are either one or zero. However, the rest is easy. Assume $r\gt s$. Computing the coefficient of $x^s$ shows that $a_{s} \in \{0,1\}$ because the coefficient of $x^s$ is the sum of $a_s$ and some other numbers that are either one or zero. By repeating this process the claim is proved for $a_{s+1}, a_{s+2},\ ...a_{r-1} .$
