We start with the quintic equation in Bring-Jerrard form f(x)=x^{5}+x+t, tεR.

We will assume that x=x(t) is a suitably smooth function of t, that x(0)=0, and that 5x^{4}+1≠0. If we then differentiate x as a function of t, via f(x) = 0, we get x'(t)=-1/(5x^{4}+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/(5x^{4}+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 x^{5}= -x-t, x^{6}= -x^{2}-tx etc. So, for example x^{9 }= x^{5}-tx^{4 }= -x-t-tx^{4 }= -tx^{4}-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 x^{5}+x+t = 0, ∆^{2 }= 4^{4}+5^{5}t^{4} .

∆^{2 }can be shown to be equal to (-1)^{10}Π_{1≤i≤5}f ‘(x_{i}).

This expression in turn can be shown to be equal to 5^{5}**Π**_{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} = 5^{5}(4/5)^{4}**Π**_{1≤i≤4 }(y_{i}+5t/4) or,

∆^{2}= 5^{5}(-4/5)^{4}**Π**_{1≤i≤4 }(-5t/4-y_{i}) .

Now f ‘(x) = 5**Π**_{1≤i≤4 }(x-y_{i}) where f ‘(y_{i})=0.

Hence Δ^{2} = 5^{5}(-4/5)^{4}(1/5)f ‘(-5t/4) or, when simplified, Δ^{2} = 4^{4} + 5^{5}t^{4}.

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

Δ^{2}x ‘ = -5(4^{3})x^{4}+5^{2}4^{2}tx^{3}-5^{3}4t^{2}x^{2}+5^{4}t^{3}x-4^{4}.

**Proof (short version)**

Assume that x≠o and 5x^{4}+1≠0.

Start with x ‘(t) = -1/(5x^{4}+1).

Then, Δ^{2}x ‘ = -(5^{5}t^{4} + 4^{4})/(5x^{4}+1)+4^{4}-4^{4}.

=-4^{4}+5((4x)^{4}-(5t)^{4})/(5x^{4}+1)

=-4^{4}+5x((4x)^{4}-(5t)^{4})/(5x^{5}+x)

=-4^{4}-5x((4x)^{4}-(5t)^{4})/(4x+5t) Note: 5x^{4}+1≠0 implies that 4x+5t≠0.

=-4^{4}-5x((4x)^{3}-(4x)^{2}5t+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 … Δ^{2}x” + 10(5^{4})t^{3}x’ = 5(4^{2})x^{3} – 3(4)5^{2}tx^{2} + 6(5^{3})t^{2}x.

Repeating this process again yields …

Δ^{2}x”’ + 30(5^{4})t^{3}x” + 135(5^{3})t^{2}x’ = -15(4^{2})x^{2} + 51(5^{2})tx.

And finally, Δ^{2}x^{iv} + 50(5^{4})t^{3}x”’ + 117(5^{4})t^{2}x” + 51(5^{4})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)!)t^{4n+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 x^{5} + 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.