Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 7 / 1

Identify the circle with $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and use normalized Fourier coefficients and convolution:For $0\le r<1$, summing two geometric series gives the Poisson kernel on the circle:This Fourier series converges absolutely and uniformly for each fixed $r<1$. Consequently we can integrate it term by term in the convolution; the substitution $u=t-s$ givesThe resulting series is itself absolutely and uniformly convergent, since $|\hat{f}(n)|\le \Vert f{\Vert }_{\infty }$ and ${\sum }_n{r}^{|n|}<\infty$.

The Poisson kernel on the circle is nonnegative and has normalized integral one, as its constant Fourier coefficient is one. Its mass concentrates near zero. Indeed, for $r\ge 1/2$ and $\delta \le |s|\le \pi$,Hence its normalized integral outside $(-\delta ,\delta )$ tends to zero as $r\uparrow 1$. Given $\eta >0$, uniform continuity of $f$ gives a $\delta >0$ such that $|f(t-s)-f(t)|<\eta$ whenever the circular distance of $s$ from zero is less than $\delta$. Splitting the convolution error into this arc and its complement yieldsThe second term tends to zero, and $\eta$ is arbitrary. Thus the Abel sums converge uniformly to $f$, or $\boxed{\Vert f\ast P_r-f{\Vert }_{\infty }\to 0}$. This is uniform Poisson summability of continuous circle functions: positivity, unit mass and concentration supply the required approximate identity argument.

Use the given integer $u$ satisfying $u^2\equiv -1(\mathrm{mod}p)$. The Euclidean lattice under consideration has basis $(1,u),(0,p)$, because every vector in it is uniquely $(a,ua+bp)$ with $a,b\in \mathbb{Z}$. Its fundamental parallelogram therefore has areaWe use the following two-dimensional form of the Minkowski convex body theorem: a convex, centrally symmetric measurable subset of ${\mathbb{R}}^2$ of area strictly greater than four times the covolume of a full-rank Euclidean lattice contains a nonzero vector of that lattice. Apply it to the closed disk of squared radius $3p/2$. Its area is $3\pi p/2>4p$, so it contains a nonzero lattice vector $(a,b)$ satisfyingOn the other hand the defining modular congruence and $u^2\equiv -1(\mathrm{mod}p)$ implyThere is exactly one positive multiple of $p$ strictly below $2p$, so $\boxed{p=a^2+b^2}$. Neither coordinate can be zero, since a prime is not the square of an integer. This congruence lattice proof of the prime sum of two squares proves the required case of the sum of two squares theorem without assuming any conclusion at the critical area $4p$.