Worked solution: Lindemann's transcendence proof of $\pi$ settles squaring the circle (1882)

The classical problem, attested since at least the 5th century BCE (Anaxagoras is said to have worked on it in prison), asks for a straightedge-and-compass construction of a square with the same area as a given circle. Normalizing the radius to $1$, the circle has area $\pi$, so the required square has side $s$ with $s^2 = \pi$, i.e. $s = \sqrt{\pi}$. The geometric task is thereby reduced to producing one segment of length $\sqrt{\pi}$ from a unit segment.

Wantzel's 1837 theorem converts 'producible segment' into algebra: every constructible length lies at the top of a tower of quadratic extensions of $\mathbb{Q}$, so it is algebraic over $\mathbb{Q}$ — a root of some nonzero rational polynomial — and its degree is a power of $2$. Hence squaring the circle is possible only if $\sqrt{\pi}$ is algebraic. Since the algebraic numbers form a field closed under polynomial operations, $\sqrt{\pi}$ is algebraic if and only if its square $\pi$ is algebraic: if $\alpha$ satisfies $P(x) = 0$, then $\alpha^2$ is algebraic because $\mathbb{Q}(\alpha^2) \subseteq \mathbb{Q}(\alpha)$ is a finite extension, and conversely $\sqrt{\pi}$ would satisfy $Q(x^2)=0$ if $\pi$ satisfied $Q(x)=0$.

So everything reduces to a question about one real number: is $\pi$ algebraic, or is it transcendental — satisfying no polynomial equation with rational coefficients at all? Legendre had conjectured irrationality of $\pi^2$ in the 1790s, and Lambert had proved $\pi$ irrational in 1761, but transcendence is far stronger: it must rule out polynomials of every degree simultaneously. Ferdinand von Lindemann settled it in 1882, extending the method Charles Hermite had used nine years earlier on $e$.