Newton's large sum

From Newton's Identities we have:

$\large \displaystyle p_k = x^k + y^k + z^k$

$\large \displaystyle e_1 = x+y+z$

$\large \displaystyle e_2 = xy+yz+xz$

$\large \displaystyle e_3 = xyz$

For this case:

$\color{#20A900} \boxed{ \large \displaystyle p_1 = 1}$

$\color{#20A900} \boxed{ \large \displaystyle p_2 = 2}$

$\color{#20A900} \boxed{ \large \displaystyle p_3 = 3}$

Calculating $e_1$ , $e_2$ and $e_3$

Since $e_1 = p_1$ :

$\color{#20A900} \boxed{ e_1 = 1}$

For $e_2$ :

$\large \displaystyle p_1^2 = 1 = x^2 + y^2 + z^2 +2(xy+yz+xz) = 2 + 2e_2$

$\color{#20A900} \boxed{ \large \displaystyle e_2 = -\frac12 }$

And for $e_3$ :

$\large \displaystyle p_1 e_2 = -\frac12 = xy^2 + x^2y + xz^2 + x^2z + yz^2 + y^2z + 3xyz$

$\large \displaystyle p_1 e_2 = -\frac12 = xy^2 + x^2y + xz^2 + x^2z + yz^2 + y^2z + 3e_3$

$\boxed{ \large \displaystyle xy^2 + x^2y + xz^2 + x^2z + yz^2 + y^2z = -\frac12 - 3e_3 }$

$\large \displaystyle p_1^3 = 1 = x^3 + y^3 + z^3 + 3(xy^2 + x^2y + xz^2 + x^2z + yz^2 + y^2z) + 6xyz$

$\large \displaystyle p_1^3 = 1 = 3 + 3 \left ( -\frac12 - 3e_3 \right ) + 6e_3$

$\color{#20A900} \boxed{ \large \displaystyle e_3 = \frac16 }$

Again from Newton's identities, for $k>3$ , we'll have:

$\large \displaystyle p_n = \sum_{i = n-3}^{n-1} (-1)^{n-1+i} e_{n-i} p_i$

$\large \displaystyle p_n = (-1)^{2n-4} p_{n-3} e_3 + (-1)^{2n-3} p_{n-2} e_2 + (-1)^{2n-2} p_{n-1} e_1$

$\large \displaystyle p_n = \frac16 p_{n-1} + \frac12 p_{n-2} + p_{n-3}$

$\large \displaystyle 6p_n = p_{n-1} + 3p_{n-2} + 6p_{n-3}$

From this point on, we have to check whether $p_{n-1} + 3p_{n-2} + 6p_{n-3}$ produces a multiple of $6$ , for $p_n$ to be an integer. Calculating the first items:

$\large \displaystyle p_4 = \frac{25}{6}$

$\large \displaystyle p_5 = 6$

$\large \displaystyle p_6 = \frac{103}{12}$

$\large \displaystyle p_7 = \frac{221}{18}$

$\large \displaystyle p_8 = \frac{1265}{72}$

From this point on each factor for $6p_n$ in the aforementioned formula will always have non-integer $6p_{n-3}$ , since the denominator of $p_{n-3}$ will always be greater than $6$ . So, $6p_n$ will never be an integer and, of course, will never be a multiple of $6$ .

So it only occurs at $n=5$ . The answer is, then, $\color{#3D99F6} \boxed{1}$