Can this be really solved?

$x^5-7x^4-1=0$ we get that $\sum (x)=7$ for roots $\frac{1}{r_i}$ , the equation becomes $y = \frac{1}{x}$ $\frac{1}{y^5}-7\frac{1}{y^4}-1=0$ $y^5+7y-1=0$ since $s_1=s_2=s_3=0$ , $p_1=p_2=p_3=0$ where $s_n$ is nth symmetric sum and $p_n$ is nth power sum. so $y^5=-7y+1--(1)$ $y^{20}=2401y^4+irrelevent+1$ irrelevent because sum of all $y,y^2,y^3$ is 0: $\sum y^{20} = 2401\sum(y^4) + 5$ $(1)--y^4=-7+\frac{1}{y}=-7+x$ $\sum y^4=-35+\sum x = -35+7=-28$ $\sum y^{20} = 2401*-28 + 5=\boxed{ -67223}$

Nice problem!!

We can rewrite the equation as

$\large{x^4=\frac{1}{x-7}}$

$x\to r_{i}$ where $i=1,2,3,4,5$

$\large{r_{i}-7=\frac{1}{r_{i}^{4}}}$

raising the power to five

$\large{\frac{1}{r_{i}^{20}}=(r_{i}-7)^5}$

the required expression is $\large{\displaystyle \sum^{5}_{i=1} \frac{1}{r_{i}^{20}}=\displaystyle \sum^{5}_{i=1} (r_{i}-7)^5}$

Expanding the expression, we get

$\large{\displaystyle \sum^{5}_{i=1} \left( r_{i}^{5}- 7 \binom{5}{1}r_{i}^{4}+7^{2}\binom{5}{2} r_{i}^{3}-7^{3} \binom{5}{3} r_{i}^{2}+7^{4} \binom{5}{4}r_{i}-7^{5}\binom{5}{5}\right) }$

now applying newtons identities, which states

$\large{P(x)={ a }_{ k }{ x }^{ k }+{ a }_{ k-1 }{ x }^{ k-1 }+.....{ a }_{ 1 }x+{ a }_{ 0 }\\ { e }_{ 1 }=\sum { { \alpha }_{ i } } =-\frac { { a }_{ k-1 } }{ { a }_{ k } } \\ { e }_{ 2 }=\displaystyle \sum { { \alpha }_{ i }{ \alpha }_{ j } } =\frac { { a }_{ k-2 } }{ { a }_{ k } } \\ { e }_{ k }=\prod { { \alpha }_{ i } } ={ \left( -1 \right) }^{ k }\frac { { a }_{ 0 } }{ { a }_{ k } } \\ { P }_{ i }=\displaystyle \sum _{ j=1 }^{ k }{ { \alpha }_{ j }^{ i } } \\ For\quad i\le k-1\\ { P }_{ k-1 }={ e }_{ 1 }{ P }_{ k-2 }-{ e }_{ 2 }{ P }_{ k-3 }+....+{ \left( -1 \right) }^{ k-2 }{ e }_{ k-1 }\left( k-1 \right) \\ For\quad i\ge k\\ { P }_{ i }=\displaystyle \sum _{ j=1 }^{ k }{ { \left( -1 \right) }^{ j+1 }{ e }_{ j }{ P }_{ i-j } } }$

Now

$\large{(\begin{matrix} { e }_{ 1 }=7 & { P }_{ 0 }=5 \\ { e }_{ 2 }=0 & { P }_{ 1 }=7 \\ { e }_{ 3 }=0 & { P }_{ 2 }={ 7 }^{ 2 } \\ { e }_{ 4 }=0 & { P }_{ 3 }={ 7 }^{ 3 } \\ { e }_{ 5 }=1 & { P }_{ 4 }={ 7 }^{ 4 } \end{matrix}\\ { P }_{ 5 }={ 7 }^{ 5 }+5}$

$\large{\left[ \because \quad { P }_{ 3 }={ e }_{ 1 }{ P }_{ 1 }-{ e }_{ 2 }{ P }_{ 1 }+{ e }_{ 3 }\times 3\\ \quad \quad { P }_{ 5 }={ e }_{ 1 }{ P }_{ 4 }-{ e }_{ 2 }{ P }_{ 3 }+{ e }_{ 3 }{ P }_{ 2 }-{ e }_{ 4 }{ P }_{ 1 }+{ e }_{ 5 }{ P }_{ 0 } \right] }$

We finally get after plugging in the values

$\large{=7^5+5\color{#D61F06}{-5\times 7^5}\color{#3D99F6}{+10\times 7^5}\color{#3D99F6}{-10\times 7^5}\color{#D61F06}{+5\times 7^5} -5\times 7^5}$

$\large{=7^5+5-5\times 7^5=\boxed{-67233}}$