Let the roots be $r_1, r_2, \dots, r_n$. By Vieta's formula, \[r_1 + r_2 + \dots r_n = \binom{n}{1} = n\] \[r_1r_2 + r_1r_3 + \dots r_1r_n + r_2r_3 + r_2r_4 + \dots r_2r_n + \dots r_{n - 1}r_n = \binom{n}{2}\] Squaring the first and subtracting twice the second, \[r_1^2 + r_2^2 + \dots r_n^2 = n^2 - 2\binom{n}{2} = n\] But by Cauchy-Schwarz, \[n = r_1^2 + r_2^2 + \dots r_n^2 \ge \frac{1}{n}(r_1 + r_2 + \dots r_n)^2 = n\] so equality holds, meaning $r_1 = r_2 = r_3 = \dots = r_n = 1$ (since their sum is $n$), and \[P(x) = (x - 1)^n\] It is easy to see that \[[x^{r}]P(x) = (-1)^{n - r}\binom{n}{r}\]

Let $\alpha_1, \alpha_2, \cdots , \alpha_n$ be the $n$ real roots of $P$. Let $Q(x) = P(x+1)$. Thus the roots of $Q$ are precisely $\beta_i = \alpha_i - 1 \ \ \forall \ 1\le i \le n$. By Binomial Theorem, it is easy to see that $Q(x)= x^n + 0x^{n-1} + 0x^{n-2} + \text{(terms of degree less than } n-2 \text{)}$. Thus by Vieta's formulae $\sum \beta_i^2 = \left(\sum \beta_i\right)^2 - \sum \beta_i\beta_j = 0$. But since $\alpha_i$ are real, so are $\beta_i$, which means that all $\beta_i$ are $0$. Thus all $\alpha_i$ are $1$ and $P(x) = (x-1)^n$.

note that , from vieta's relation ,the sum of n roots=$n$ the sum of roots 2 taken at a time=$n(n-1)/2$ so , clearly ,sum of squares of the roots=$n$ let the roots be $a_{1},...,a_{n}$ so ,$n\sum{a_{i}^2}=(\sum{a_{i})^2}$. the roots are all real .so , by the equality condition of CS inequality , we have that , all the roots are equal and trivially , all of them are 1. so ,the rest is trivial now

mathbuzz wrote: note that , from vieta's relation ,the sum of n roots=$n$ the sum of roots 2 taken at a time=$n(n-1)/2$ so , clearly ,sum of squares of the roots=$n$ let the roots be $a_{1},...,a_{n}$ so ,$n\sum{a_{i}^2}=(\sum{a_{i})^2}$. the roots are all real .so , by the equality condition of CS inequality , we have that , all the roots are equal and trivially , all of them are 1. so ,the rest is trivial now I didn't understand if sum of roots is n then how come sum of squares of the roots is n

@above see carefully. It's $n\sum a_i^2=\left(\sum a_i\right)^2$

I understand

Equality holds in cauchy schwarz, hence all the roots must be 1

Quite similar to above, but still posting, Let the roots be $v_1,v_2,v_3...v_n$, By Vieta's formula, \[v_1 + v_2 + \dots v_n = \binom{n}{1} = n\]\[v_1v_2 + v_1v_3 + \dots v_{n-1}v_n = \binom{n}{2} \]SO, \[v_1^2+v_2^2+ \dots v_n^2 = n\]By RMS - AM We have \[v_1=v_2=v_3= \dots = v_n =1 \]So \[a_k = (-1)^k \binom{n}{k}\]