Solving quintic equations of the form $x^5-x+A=0$ I was on Wolfram Alpha exploring quintic equations that were unsolvable using radicals.  Specifically, I was looking at quintics of the form $x^5-x+A=0$ for nonzero integers $A$.  I noticed that the roots were always expressible as sums of generalized hypergeometric functions:
$$B_1(_4F_3(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{3125|A|^4}{256}))+B_2(_4F_3(\frac{7}{10},\frac{9}{10},\frac{11}{10},\frac{13}{10};\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3125|A|^4}{256}))+B_3(_4F_3(\frac{9}{20},\frac{13}{20},\frac{17}{20},\frac{21}{20};\frac{3}{4},\frac{5}{4},\frac{3}{2};\frac{3125|A|^4}{256}))+B_4(_4F_3(\frac{-1}{20},\frac{3}{20},\frac{7}{20},\frac{11}{20};\frac{1}{4},\frac{1}{2},\frac{3}{4};\frac{3125|A|^4}{256}))$$
where the five roots have $(B_1,B_2,B_3,B_4)\in\{(A,0,0,0),(-\frac{A}{4},-\frac{5A|A|}{32},\frac{5|A|^3}{32},-1),(-\frac{A}{4},\frac{5A|A|}{32},-i\frac{5|A|^3}{32},i),(-\frac{A}{4},\frac{5A|A|}{32},i\frac{5|A|^3}{32},-i),(-\frac{A}{4},-\frac{5A|A|}{32},-\frac{5|A|^3}{32},1)\}$
After observing this, I was left with a lot of questions.   First, given $A$, is there a formula I can use to generate the values for $D$, $E$, $H$, and $K$?  Second, why do these patterns persist?  Third, if I take a different set of quintics that can't be solved using radicals and that differ only in their constant term, does a similar pattern to the roots exist?  Fourth, can anyone prove that these patterns that I found will always hold?
Edit: I found the patterns for $D$, $E$, $H$ and $K$.  The question has been updated accordingly.
 A: The answer to your third question is yes! The method uses Bring radicals, whose explicit form in terms of generalized hypergeometric functions can be found using the Lagrange inversion theorem. (In fact since any quintic can be reduced to this form, in principle this method can be used to solve any quintic.) I can answer your second and fourth questions partially by developing this method. But I'm afraid I will only be able to obtain the first solution with coefficients $(A, 0, 0, 0)$.
The idea at its core is quite simple. Basically we rewrite the equation as $x^5 - x = - A$, treat the left hand side as a function $f(x) = x^5 - x$, then try to answer the question "what is $f^{-1}(-A)$." This is then done by expressing $f^{-1}$ as a power series. The Lagrange inversion theorem gives this inverse as
$$
x = \sum_{k=0}^\infty \binom{5 k}{k} \frac{A^{4k+1}}{4k+1}\ .
$$
Unfortunately, this series doesn't converge for all values of $A$. In fact the radius of convergence is $4/(5\times 5^{1/4})\approx 0.535$, so evaluating the series directly would only give us the solution for one integer $A = 0$. This is where the generalized hypergeometric function comes in. We can analytically continue this series to define a function of $A$. The function whose power series (at zero) is
$$
\sum_{n=0}^{\infty}\prod_{k=0}^{n} \frac{(k+a_1)\cdots(k+a_p)}{(k+b_1)\dots(k+b_q)(k+1)} z
$$
is denoted as $_p F_q(a_1,\dots, a_p;b_1,\dots,b_q;z)$. To convert our function $f^{-1}(A)$ into the standard form, we need to compute the ratio between consecutive terms, which is
$$
\begin{align}
& \quad \frac{(5k +5)!A^{4k+5}}{(k+1)!(4k+4)!(4k+5)}\cdot\frac{k!(4k)!(4k+1)}{(5k)!A^{4k+1}}\\
& = \frac{(5k+5)(5k+4)(5k+3)(5k+2)(5k+1)(4k+1)A^4}{(k+1)(4k+4)(4k+3)(4k+2)(4k+1)(4k+5)} \\
& = \frac{5(5k+4)(5k+3)(5k+2)(5k+1)}{4(4k+5)(4k+3)(4k+2)(k+1)}A^4 \\
& = \frac{(k+1/5)(k+2/5)(k+3/5)(k+4/5)}{(k+1/2)(k+3/4)(k+5/4)(k+1)}\left(5\left(\frac{5A}{4}\right)^4\right)\ .
\end{align}
$$
Now since the numerator has four factors and the denominator has three factors besides $(k+1)$, this is $_4F_3$ (times an extra factor of $A$, since the starting term in our series is $A$, not $1$). The parameters are the numbers added to $k$ in each factor, and the argument is $(5^5/4^4)A^4 = (3125/256)A^4$. This gives you the first solution $A \;_3F_4(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{3125}{256}A^4)$.
In order to obtain the other roots, in principle we can find use this root to factor the polynomial and try to solve the resulting quartic. However that involves too much computation and doesn't seem like the neatest way of obtaining the results you got here.
