Find all polynomials whose coefficients are $\{0,\ldots, n\}$ and all roots are real Find all polynomials $P \in \mathbb{Z}[x],\;\deg(P)= n$ s.t.:
$(1)$ All roots of $P$ are real
$(2)$ The set of coefficients of polynomial $P$ is equal to $\{0, 1, ..., n\}$
Due to Vieta's formula, I have realized that the coefficient $a_0$ (the one that is with $x^0$) must be equal to $0$.
 A: Notice that every polynomial satisfying the two conditions must be positive on $\mathbb{R}$, therefore all its roots are negative. 
Now Let's write $P=X\prod_{1\leq k \leq n-1}(X+a_{i})$ 
we have $(\sum_{1 \leq k \leq n-1} a_{k})((\sum_{1 \leq k \leq n-1} \frac1{a_{k}})\geq (n-1)^2$, therefore $(\sum_{1 \leq k \leq n-1} a_{k})((\sum_{1 \leq k \leq n-1} (\prod_{1 \leq j \leq n-1} a_{j})\frac1{a_{k}})\geq (n-1)^2(\prod_{1 \leq j \leq n-1} a_{j})$
Notice that the coefficient of $x^{n-1} $(respectively $x$) in the expansion of $P=\prod_{1\leq k \leq n-1}(x+a_{i})$
is $(\sum_{1 \leq k \leq n-1} a_{k})$ (respectively $(\sum_{1 \leq k \leq n-1} (\prod_{1 \leq j \leq n-1} a_{j})\frac1{a_{k}})$ ). therefore there product is inferior to $n(n-1)$
but this implies $ \frac{n}{n-1}\geq (\prod_{1 \leq j \leq n-1} a_{j}) $ 
therefore being the coefficient of $x^{1}$ in $P$ the quantity  $ (\prod_{1 \leq j \leq n-1} a_{j}) =1 or 2$. it equals 2 if and only if n=2 i.e $P=X(X+2)$ .
I think that if you push the reasoning and consider the coefficient of $x^{k}$ times the coefficient of $x^{n-k}$ you can prove that this is the only solution for $n\geq2$.  
