The problem can be solved with a simple idea that needs more efforts to be written down than to be explained. b) We will construct ${ P_i(x) = \prod_{j=1}^{n} (x-x^{(i)}_j}) $, where $x^{(i)}_j $ are integers and are all different. A construction 1. Let arbitrary choose different integers $x^{(1)}_j,\,j=1,2,\ldots,n$. 2. Suppose $\{x^{(i)} \}_{j=1}^{n},\,i=1,2,\ldots,m $ are determined and let $ x^{(i)}_j < x^{(i)}_{j+1}$. 3. Denote $M(A, B) = \sum_{i=1}^m \,\, \sup_{ x \in [A,\, B] } |P(x)|$. Now choose different integers $ \{x^{(m+1)} \}_{j=1}^n$, that are also different from $x^{(i)}_j, i=1,2,\ldots,m,\, j=1,2,\ldots,n$ and such that ${ P_{m+1}(x) = \prod_{j=1}^{n} (x-x^{(m+1)}_j}) $ satisfies the following properties: $ |P_{m+1}(x^{(m+1)}_1 - 2|x^{(m+1)}_1| )| > M(x^{(m+1)}_1 - 2|x^{(m+1)}_1|, x^{(m+1)}_n + 2|x^{(m+1)}_n|)$ $ |P_{m+1}(\frac{x^{(m+1)}_j + x^{(m+1)}_{j+1}}{2})| > M(x^{(m+1)}_1 - 2|x^{(m+1)}_1|, x^{(m+1)}_n + 2|x^{(m+1)}_n|) ,\, j=1,2,\ldots,n-1 $ $ |P_{m+1}(x^{(m+1)}_n + 2|x^{(m+1)}_n|)| > M(x^{(m+1)}_1 - 2|x^{(m+1)}_1|, x^{(m+1)}_n + 2|x^{(m+1)}_n|) $ I skip the details but it is easy to see that such $ \{x^{(m+1)} \}_{j=1}^n$ can be chosen. 4. Go to step 2. $\square $ Now let $P_{i_1},P_{i_2},\ldots, P_{i_s}$ are some of these polynomials. Let $i$ is the biggest index amongst ${i_1,i_2,\ldots, i_s}$. Then the construction ensures that $\sum_{j=1}^s P_{i_j}(x) $ has alternating signs in the points $x^{(i)}_1 - 2|x^{(i)}_1|,\frac{x^{(i)}_1 +x^{(i)}_2}{2},\ldots , \frac{ x^{(i)}_{n-1} + x^{(i)}_n }{2}, x^{(i)}_n + 2|x^{(i)}_n| $. So $\sum_{j=1}^s P_{i_j}(x) $ has n different real roots.

A little bit wrong in the technical details, though it can be fixed. There is no need $P_{m+1}$ in the "alternating" knots to dominate the sum of the norms of $P_i , \, i \leq m $ over the whole interval. It is too strong and could not be achieved. It is enough $P_{m+1}$ in the "alternating" knots to dominate just the sum of the values of $P_i , \, i \leq m $ in these "knots". And You can construct $x^{(i)}_j$ as if going away fast from one another just as the galaxies do!

let $ b_{0} > b_{1}> ...> b_{n}$ be positive integers and $\delta _{i,j}$ is a negative integer if $j$ is odd and $\delta _{i,j}$ is a positive integer if $j$ is even and . and for all $1\leq i\leq k,0\leq j\leq n$ we have $(\prod_{i=1}^k b_{i})r_{s,j}=\delta _{s,j}$ such that $\sum_{s\geq 0}^n r_{t,s}=1$. let $v_{i}(x)=\prod _{j=0}^n(x-b_j)/(x-1)(x-i)$.We define $P_{1}(x)=\sum r_{1,s}\prod _{j\geq 0,j\neq 1}(x-b_{j})$we put $r_{1,1}=r_{1,2}=t$. and $p_{i}(x)=-l_{i}tv_{i}(x)(x-b_{i})+ltv_{i}(x)(x-b_{1})+p_{1}(x)$. and we can get the $p_{i}(x)$ be $\sum r_{i,f}\prod _{j\geq 0,j\neq i}(x-b_{j})$. by Lagrange Formula we can see $p_{i}(b_{q})=\delta _{i,q}$ for all $0\leq q\leq n$. so we have $F(x)=\sum P_{c_{d}}(x)$ insert the $x$ exists exactly $n+1$ time so has exactly $n$ real roots since $F(x)$has at least one real root in each of these intervals $(b_{0},b_{1}),(b_{1},b_{2}),...,(b_{n-1},b_{n})$ ( for all the $P_{c_{d}}(x)$ which has $n$ degree and is a monic polynomial ). by our definition for these polynomials you can see $(P_{z}(x),p_{w}(x))=1$ . edit --> $x-b_m$ dose not divide the polynomial $\prod _{j\geq 0,j\neq m}(x-b_{j})$.

I think that we can use the same idea as in this problem, i.e. we can put $P_n(x)=x^n-c_n(x-1)(x-2)\cdots (x-(n-1))$ where $c_n$ is very big number and then use intermediate value theorem to prove that all these polynomials have all real roots.

Yes I use this idea in my solution .but we should be careful aboute $ A=\{P_1(x),.....,P_k(x)\} $ such that no two of them have a common factor.