Find all positive integers $n$ ,such that the polynomial $ a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n $is irreducible over the real numbers 
Find all positive integers $n$ with $n \ge 2$ such that the polynomial $$P(a_1, a_2, ..., a_n) = a_1^n+a_2^n + ... + a_n^n - n a_1 a_2 ... a_n $$ in the $n$ variables $a_1$, $a_2$, $\dots$, $a_n$ is irreducible over the real numbers, i.e. it cannot be factored as the product of two nonconstant polynomials with real coefficients.

Note that $ x^2 - 2xy + y^2 = (x - y)^2 $ and that $ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $ so the relevant polynomial is factorable for $ n = 2 $ and $ n = 3 $
Now I  conjecture for $ n \geq 4 $ this polynomial is irreducible. But How prove?
 A: Short answer : Your conjecture is true, and the polynomial is in fact irreducible over
$\mathbb C$. This is because your polynomial is basically a cyclotomic polynomial plus
a negligible pertubation. Your polynomial is so close to a cyclotomic one that if it had
any nontrivial divisor, it would have to be a "cyclotomic" divisor, but the perturbation prevents that.
Detailed answer :
Let  $P(a_1,a_2,\ldots,a_n)=\sum_{k=1}^n a_k^n-na_1a_2\ldots a_n$ be your polynomial, and let $P^{\tilde{}}(X,Y)=P(1,1,\ldots,1,X,Y)$. It will suffice to show that  $P^{\tilde{}}$ is irreducible in ${\mathbb C}[X,Y]$ (indeed, suppose that $P=A\times B$ is a nontrivial factorization of $P$. Then $P^{\tilde{}}=A^{\tilde{}} \times B^{\tilde{}}$, so one of the two factors (say $A^{\tilde{}}$) is constant in  ${\mathbb C}[X,Y]$. Then $B^{\tilde{}}$ has maximal degree $n$ in both $X$ and $Y$, which implies the same for $B$. So $A$ has degree $0$ in both $X$ and $Y$. Then $A$ must divide all the $(X,Y)$-coefficients in $P$, namely $\sum_{k=1}^{n-2} a_k^n, -na_1a_2\ldots a_{n-2},1$ and $1$. So $A$ must be constant).
By some asymptotic root analysis (see appendix), there are constants 
$M,c>0$ such that for any $x\in [M,\infty)$, the polynomial $P^{\tilde{}}(x,Y)$
has roots $\alpha_1(x),\ldots,\alpha_n(x)$ with 
$$|\alpha_k(x)-\zeta_k x|\leq\frac{c}{x^{n-3}}\tag{1}$$
for each $k$, where $\zeta_k=e^{\frac{k}{n}i\pi}$. 
Denote by $s_k$ the $k$-th elementary symmetric polynomial,
$s_k(u_1,\ldots,u_n)=\sum_{i_1<\ldots i_k}u_{i_1}\ldots u_{i_k}$. And for 
$I\subseteq \lbrace 1,2,\ldots, n\rbrace$, say $I=\lbrace i_1,i_2,\ldots i_r \rbrace$ put $s_I(u_1,\ldots,u_n)=s_{r}(u_{i_1},u_{i_2},\ldots,u_{i_r})$. Finally let
$\alpha(x)=(\alpha_1(x),\ldots,\alpha_n(x))$ and $\zeta=(\zeta_1,\ldots,\zeta_n)$. By multiplying and adding, we have for any $I\subseteq \lbrace 1,2,\ldots, n\rbrace$,
a constant $c_I>0$ such that
$$
|s_I(\alpha(x))-s_I(\zeta)x^{|I|}|\leq \frac{c_I}{x^{n-(|I|+2)}}  \tag{2}
$$
Suppose that there is a nontrivial factorization $P^{\tilde{}}=A \times B$ in ${\mathbb C}[X,Y]$. We may suppose that $A$ and $B$ are monic in $Y$.  Note that of the two factors (say $A$) has degree $\leq n-3$ in $Y$, otherwise both factors have degree at least
$n-2$, whence $n\geq 2(n-2)$ which is false for $n\geq 5$ (the case $n=4$ is similar and simpler).
Let $r$ be the degree of $A$ in $Y$. For any $x$, $A(x,Y)=\prod_{i\in I_x}(Y-\alpha_i(x))$ for some $I_x\subseteq \lbrace 1,2,\ldots, n\rbrace$ with cardinality $r$. Since there are only finitely many possible values for $I_x$, by the pigeon-hole principle there is an infinite $X\subseteq [M,\infty)$ and a fixed $I$ such that $I_x=I$ for all $x\in X$. Then for any $J\subseteq I$, $s_J(\alpha(x))$ is a coefficient appearing in $A(x,Y)$, so it must be a polynomial in $x$. Then (2) (used with $J$ in place of $I$) forces this polynomial to be $s_J(\zeta)x^{|J|}$. It follows that for any $x\in X$, $A(x,Y)=
\prod_{i\in I}(Y-\zeta_ix)$. This identity holds for any $x$ because $X$ is infinite. So $A$ divides $X^n+Y^n$. But then $A$ also divides $P^{\tilde{}}-(X^n+Y^n)=-nXY+(n-2)$, which forces $r=1, A=Y-\zeta_kX$ for some $k$, and is clearly impossible. This finishes the proof.
Appendix : asymptotic root analysis 
Here when we say "constant" we mean "independent of $x$". Let 
$$ Q_x(T)=P^{\tilde{}}(x,xT)=x^n(T^n+1)-nx^2T+n-2 $$ 
Step 1 : There are two constants $K_1,M_1>0$ such that $|t| \leq K_1$ whenever 
$t$ is a root of $Q_x$ and $x\in (M_1,\infty)$.  
It suffices to take $K_1,M_1$ such that $M_1\geq 1,K_1^{n-1}\geq n, K_1^n\geq 6(n-2)$. 
Indeed, if $|t|>K_1$ and $x\in (M_1,\infty)$, we have $|x^nt^{n-1}| \geq x^n K_1^{n-1}$ whence $|x^nt^{n-1}-nx^2| \geq  x^n K_1^{n-1} -nx^2 \geq x^n\frac{K_1^{n-1}}{2}$
and $|x^nt^n-nx^2t|\geq x^n\frac{K_1^n}{2}$, and eventually
$|Q_x(t)| \geq x^n\frac{K_1^n}{2}-((n-2)+x^n) \geq x^n\frac{K_1^n}{6}$, so $t$ cannot
be a root of $Q_x$.
Step 2 : There are two constants $K_2,M_2>0$ such that $|t^n+1| \leq \frac{K_1}{x^{n-2}}$ whenever $t$ is a root of $Q_x$ and $x\in (M_2,\infty)$.  
Thanks to step 1 and the fact that $t^n+1=\frac{nx^2t+n-2}{x^n}$, we see that
we can take $K_2=n(K_1+1),M_2=M_1$.
The inequality in step 2 can be rewritten $\prod_{k=1}^n |t-\zeta_k| \leq \varepsilon$
where $\varepsilon =\frac{K_1}{x^{n-2}}$. In particular, $t$ satisfies at least one of the
$n$ inequalities $|t-\zeta_k| \leq \varepsilon^{\frac{1}{n}}$. For large enough $x$,
those inequalities will be mutually exclusive, showing
Step 3 : There are two constants $K_3,M_3>0$ such that whenever $x\in (M_3,\infty)$, $Q_x$ has roots $t_1,\ldots,t_n$ such that $|t_k-\zeta_k| \leq \frac{K_3}{x^{\frac{n-2}{n}}}$.
Putting $\varepsilon=t_k-\zeta_k$, we then have $\varepsilon \to 0$ as $x\to \infty$. Let
$C=\frac{(\zeta+\varepsilon)^n-\zeta^n-n\zeta^{n-1}\varepsilon}{\varepsilon^2}$. By expanding the Newton binomial, we have that $C \to \frac{n(n-1)}{2}\zeta^{n-2}$ when
$x\to \infty$. Now, note that $0=Q_x(t)=Q_x(\zeta_k+\varepsilon)$ can be rewritten
as 
$$
\varepsilon=\frac{nx^2(\zeta+\varepsilon)-(n-2)+C\varepsilon^2}{n\zeta_k^{n-1}x^n}
$$
We deduce
Step 4 : There are two constants $K_4,M_4>0$ such that whenever $x\in (M_4,\infty)$, $Q_x$ has roots $t_1,\ldots,t_n$ such that $|t_k-\zeta_k| \leq \frac{K_4}{x^{n-2}}$.
Multiplying by $x$ in step 4, we obtain (1) as announced.
A: This problem has a solution https://artofproblemsolving.com/community/c6h599342p3557426.
