Minimising an expression - involving polynomial I found this one on a forum but it has been unanswered from long there. I am curious to know if there is a solution to this problem. Here it is:
Let n be a positive integer. Determine the smallest possible value of
$$|p(1)|^2+|p(2)|^2 + .........+ |p(n+3)|^2$$
over all monic polynomials p with degree n.

I don't have much idea about the problem. I can go about by assuming $p$ to be some monic polynomial and find $p(1)$, $p(2)$.... in terms of the coefficients but I don't think that is going to help. This is some inequality related problem, I guess.
Any help is appreciated. Thanks!
PS: Can somebody please check if I have added the proper tags? I am not sure where this kind of problem should belong. 
 A: Let us define, for two polynomials $p,q$ of degree at most $n$, the bilinear form 
$$(p,q):=\sum_{k=1}^{n+3}p(k)q(k)$$
Observe that this bilinear form is a scalar product since $||p||^2:=(p,p)$ is positive definite. $$||p||^2=0\implies p=0$$
We need to find the closest point of $A:=\{p:\ p\text{ is monic}\}$ to the origin. Equivalently we can find the distance of $-x^n$ to the linear subspace $B:=A-\{x^n\}=\{p:\ \text{deg}(p)<n\}$.
We know that the polynomial providing the minimum distance is the orthogonal projection $P(x)$ of $v:=v(x):=-x^n$ onto $B$. If $v_1(x),v_2(x),...,v_n(x)$ is an orthonormal basis of $B$ then 

$$P(x)=\sum_{k=1}^{n}(-x^n,v_k)v_k(x)\ \ \ \ \ \ \ \ \ \  \ \text{(1)}$$

To obtain an orthonormal basis of $B$ we can apply Gram-Schmidt algorithm to a basis of $B$, say $1,x,..,x^{n-1}$.
To make the computation let us denote $$s_m:=\sum_{k=1}^{n+3}k^m$$
We see that $$(x^r,x^s)=s_{r+s}$$
We can write the result of Gram-Schmidt in determinant form as 
$$v_i(x)=\frac{1}{\sqrt{D_{j-i}D_j}}\text{det}\begin{bmatrix}s_0&s_1&...&s_{j-1}\\s_1&s_2&...&s_j\\\vdots&\vdots&\ddots&\vdots\\s_{j-2}&s_{j-1}&...&s_{2j-3}\\1&x&...&x^{j-1}\end{bmatrix}$$
where
$$D_j=\text{det}\begin{bmatrix}s_0&s_1&...&s_{j-1}\\s_1&s_2&...&s_j\\\vdots&\vdots&\ddots&\vdots\\s_{j-1}&s_{j}&...&s_{2j-2}\end{bmatrix}$$
We can now compute 
$$(-x^n,v_k(x))=\frac{1}{\sqrt{D_{j-i}D_j}}\text{det}\begin{bmatrix}s_0&s_1&...&s_{j-1}\\s_1&s_2&...&s_j\\\vdots&\vdots&\ddots&\vdots\\s_{j-2}&s_{j-1}&...&s_{2j-3}\\-s_n&-s_{n+1}&...&-s_{n+j-1}\end{bmatrix}$$
So, the polynomial that minimizes the sum is 

$$x^n+P(x)$$

A: Cross-posted and answered 
on Mathoverflow. 
The formula is $(2n+1)(2n+3) n!^4 / (2n)!$.
See my answer to 
MO_210823 for the proof.
In general the minimum value of $\sum_{i=1}^{n+k+1} p(i)^2$
over monic polynomials $p$ of degree $n$ is
$$
\frac{n!^4}{(2n)!} {2n+1+k \, \choose k},
$$
and the minimizing $p$ is proportional to a
Hahn polynomial
$Q_n$ evaluated at $x-1$ (with $\alpha=\beta=0$ and $N = n+k+1$).
