Given *Ramanujan’s constant*,

where , why do we know, *in advance*, that the quintic,

is solvable in radicals? The answer is this: The general quintic can be transformed in radicals to the one-parameter form,

called the *Brioschi quintic*. Whether reducible or not, if it is solvable in radicals and *c* is rational, then it can be shown *c* must have the form,

where,

for some radical *u*. (For example, *u* = 1 will yield an *irreducible* though solvable quintic.) But it seems Nature likes to recycle polynomials as this is also one of the many formulas for the *j-function* , namely,

where,

and is the *Dedekind eta function*. However, if we let,

then we get the more well-known j-function formula,

where the numerator and denominator are polynomial invariants of the *icosahedron*,

a Platonic solid wherein one can find *pentagons* (which, of course, has 5 sides). Perhaps not surprisingly, *r* is given by the eta quotient,

But one can also express *r* using the beautiful *Rogers-Ramanujan continued fraction*. Let,

then,

One of the simplest cases is,

which was communicated by Ramanujan to Hardy in his famous letter.

Interesting, isn’t it?

For a related topic, kindly read “*Ramanujan’s Continued Fractions and the Platonic Solids*“.