Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of equation
$$
x^5-\binom{p}{1}x^4+\binom{p}{2}x^3-\binom{p}{3}x^2 + \binom{p}{4}x - \binom{p}{5} = 0.
\tag{2}
$$
I want to create related task for my son: to find polynomial $(2)$ for given system $(1)$. But it would be great to create task, where all $a,b,c,d,e$ are real and distinct. To have possibility to check manually all  the sums (without complex numbers).
Question: Which values of $p$ provide $5$ real (pairwise) distinct $a,b,c,d,e$? And if there exists such one at all (for $5$ variables)?

Using the mean value theorem repeatedly (by taking 3 derivatives), we get that $$60x^2-4 \dbinom{p}{1}x+6 \dbinom{p}{2}$$ has to have $2$ real roots. This only happens if $$\Delta_x > 0 \iff 0 < p < \frac{60}{59}.$$
Using Geogebra, it seems that none of the values in this range produce a function with $5$ real roots so we can be pretty confident in saying that no value of $p$ satisfies what you are looking for. I think we can use the fact that the quitic has a horizontal tangent at $x = 0$ for $p = 0,1$ to prove this observation but it seems messy.
