We start with the quintic equation in Bring-Jerrard form f(x)=x⁵+x+t, tεR.

We will assume that x=x(t) is a suitably smooth function of t, that x(0)=0, and that 5x⁴+1≠0. If we then differentiate x as a function of t, via f(x) = 0, we get x'(t)=-1/(5x⁴+1). Clearly x(0) = 0 implies that x'(0)=-1.

Further differentiation of x(t), with evaluation at t = 0, shows that x”(0) = 0 and x”'(0) = 0. These values will provide the initial conditions for the solution of a 4th order LDE satisfied by x(t).

The next stage is to write/expand the expression -1/(5x⁴+1)as a power series in x and use the fact that for powers of x which are 5 or higher these can be expressed as linear combinations of x to the power 0, 1, 2, 3, or 4. This follows from the fact that x⁵= -x-t, x⁶= -x²-tx etc. So, for example x⁹ = x⁵-tx⁴ = -x-t-tx⁴ = -tx⁴-x-t.

The discriminant ∆² appears in the following calculations. ∆² =Π_1≤i<j≤5 (x_i – x_j)², where x_i and x_j are pairs of distinct roots of f(x)=0. For the quintic x⁵+x+t = 0, ∆² = 4⁴+5⁵t⁴ .

∆² can be shown to be equal to (-1)¹⁰Π_1≤i≤5f ‘(x_i).

This expression in turn can be shown to be equal to 5⁵Π_1≤i≤4 f (y_i) where f ‘(y_i)=0.

It is easily shown that f(y_i)=(4/5)(y_i+5t/4), and so ∆² = 5⁵(4/5)⁴Π_1≤i≤4 (y_i+5t/4) or,

∆²= 5⁵(-4/5)⁴Π_1≤i≤4 (-5t/4-y_i) .

Now f ‘(x) = 5Π_1≤i≤4 (x-y_i) where f ‘(y_i)=0.

Hence Δ² = 5⁵(-4/5)⁴(1/5)f ‘(-5t/4) or, when simplified, Δ² = 4⁴ + 5⁵t⁴.

The details of the calculation are tedious to write out but here is the resulting expression …

Δ²x ‘ = -5(4³)x⁴+5²4²tx³-5³4t²x²+5⁴t³x-4⁴.

Proof (short version)

Assume that x≠o and 5x⁴+1≠0.

Start with x ‘(t) = -1/(5x⁴+1).

Then, Δ²x ‘ = -(5⁵t⁴ + 4⁴)/(5x⁴+1)+4⁴-4⁴.

=-4⁴+5((4x)⁴-(5t)⁴)/(5x⁴+1)

=-4⁴+5x((4x)⁴-(5t)⁴)/(5x⁵+x)

=-4⁴-5x((4x)⁴-(5t)⁴)/(4x+5t) Note: 5x⁴+1≠0 implies that 4x+5t≠0.

=-4⁴-5x((4x)³-(4x)²5t+4x(5t)²-(5t)³).

If we differentiate both sides of this equation, then after some rearrangement we end up with an expression where the left-hand side is a linear combination of x” and x’, while the right hand side has no term in x where the power of x is greater than 3.

This is … Δ²x” + 10(5⁴)t³x’ = 5(4²)x³ – 3(4)5²tx² + 6(5³)t²x.

Repeating this process again yields …

Δ²x”’ + 30(5⁴)t³x” + 135(5³)t²x’ = -15(4²)x² + 51(5²)tx.

And finally, Δ²x^iv + 50(5⁴)t³x”’ + 117(5⁴)t²x” + 51(5⁴)tx’ – 1155x = 0.

This equation can be solved using standard power series methods, with the initial conditions above, i.e. x(0)=0, x'(0)=-1, x”(0)=0, x”'(0)=0, to give …

x =∑_0≤n<∞ (-1)ⁿ⁺¹ (5n)!/(n!(4n+1)!)t⁴ⁿ⁺¹ ,|t|<4/(5 ⁴√5).

The initial condition(s) for the other roots come from the non-zero complex roots of x(0)⁴ = -1,

i.e. x(0)=(1+i)/√2, … , -(1+i)/√2.

The existence and uniqueness theorems for ODE+LDE’s ensure that this solution to the above LDE is also a solution to

the quintic equation x⁵ + x + t = 0, tεR.

A method of solving polynomial equations in this style was first discovered in the mid 19th century by Cockle and Harley.