Thread by @XorNinja

Kepler: so, you see, orbit of a planet is elliptical. To find where the Earth is, we need a method to calculate the arc length of the ellipse
Fermat: I have discovered such a marvelous method which this tweet is too narrow to contain

Newton: This is a fluent problem, and, as usual, it can be solved with an infinite series expansion
Leibniz: Pardon monsieur, what's a fluent? Arc length is an integral problem, and to calculate ∫f(x)dx first you need to find a closed form function such that F'(x) = f(x)

(100 years had passed. The search for Leibniz's closed-form solution for the elliptic integral, that is ∫f(x, g(x))dx where f is a rational function and g is a polynomial of degree 3 or 4, had been fruitless)

(By the end of the 17th century, Bernoulli -- which one is left as an exercise -- conjectured that the task is impossible. It was finally confirmed by Liouville in the 19th century who proved that elliptic integrals, as many others, are non-elementary)

(In the mean time, Fagnano found a double law and later Euler discovered general addition laws for the elliptic integral. Elliptic, double, addition! You can see where I'm heading to ;-)

(Legendre systematically reduced elliptic integrals to just three kinds, with their various addition and transformation laws. Poor Legendre, what happened next made much of his life's work obsolete as soon as it was published)

Abel: ∫1/sqrt(1 - t^4)dt looks similar to arcsine(x) = ∫1/sqrt(1 - t^2)dt, but sine is more interesting than arcsine, isn't it? Well, then, why don't we invert elliptic integrals?
Jacobi: woohoo, elliptic functions ftw!
Gauss: the fact is I found the inverses before Herr Abel.

Eisenstein (not that Einstein): all elliptic functions of a special form must satisfy y^2 = p(x), where p is a cubic polynomial with no repeated roots.

(A cubic polynomial with no repeated roots?! It's an elliptic curve!)

Weierstrass: all your elliptic function are belong to mine, and mine satisfies y^2 = 4 * x^3 − a * x^2 − b
Poincaré: the points on this curve form a group structure. Is it finitely generated?
Mordell: yes, very well monsieur.

Taniyama-Shimura: all rational elliptic curves are modular. Well, maybe, maybe not.
Frey: if a, b, c are solutions to x^n + y^n = z^n, then y^2 = x(x - a^n)(x - b^n) looks very weird
Serr - Ribet: weird or not, we don't know yet, but it wouldn't be modular

Wiles - Taylor: it is indeed weird, in fact it does not exist, because such and such elliptic curves must be modular! Fermat's last theorem QED.

In a parallel development:
Diffie-Hellman: give us a group we shall encrypt the world
Lenstra: okay, just don't encrypt with RSA. I can factor the modulus, with elliptic curves!
Miller: isn't there a group in elliptic curves? Let's encrypt!
Koblitz: of course.