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)?
 A: In order for the roots to transition from real to complex, they must pass through a double-root first - that is, the gradient must be zero at those points. By examining how many roots are present for one solution between each pair, we can determine the regions for which all solutions are real.
$$
5x^4-4\binom{p}{1}x^3+3\binom{p}{2}x^2-2\binom{p}{3}x + \binom{p}{4}= 0
$$
Since we need roots equal to extrema, we can examine the resultant of this polynomial with the original one, which gives us (evaluated using Maxima)
$$
955514880\,{\left( p-5\right) }^{4}\,{\left( p-4\right) }^{3}\,{\left( p-3\right) }^{3}\,{\left( p-2\right) }^{3}\,{\left( p-1\right) }^{3}\,{p}^{4}
$$
and this must equal zero for a root to be a double root (or higher), and thus this occurs at $p\in\{0,1,2,3,4,5\}$.
Now, we need only determine the number of roots for $p<0$, $0<p<1$, and so on. The easiest way to do this is to test $p=\frac{2n-1}2$ for $n\in\{0,1,2,3,4,5,6\}$. For $n=0$, we have one root (near -0.7). For $n=1$, we have one root (near 0.85). For $n=2$, we have three roots (near -0.25, 0.65, and 1.05). For $n=3$, we have one root (about 0.5). For $n=4$, we have three roots (near -0.05, 0.33, and 1.25). For $n=5$, we have one root (near 0.15). And for $n=6$, we have one root (near 1.75).
Of course, we could also check the integer $p$ values... but since we already know that $p\in\{0,1,2,3,4,5\}$ produces multi-roots, and thus the roots aren't distinct (as required), we need not consider those cases.
And therefore, it will never have five distinct, real solutions. The best it achieves in terms of distinct roots is three, which occurs when $1<p<2$, and when $3<p<4$.
A: One can also use inequalities to give some restrictions on the solution. 
We have $$ \sum a^2(a-1)^2 = \sum a^2 - 2 \sum a^3 + \sum a^4 = p - 2p + p = 0 $$ any solution $(a,b,c,d,e)$. Since $x^2(x-1)^2 \ge 0$ for all real $x$, each of $a, b, c, d, e$ must be either $0$ or $1$. Obviously, they cannot be all distinct. 
A: 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.
