Minkowski thm: If C is convex, symmetric to the origin in R^d and vol(C)>2^d, then it contains a gridpoint.
Examples why all three conditions are necessary.
Proof: We can suppose that C is bounded. Take n^d translates of C/2 (with nxnx..n gridpoints). They are all contained in (n+2b)^d<n^d(1+eps)^d= sum of vol of translates, so two translates intersect. Using a translation, we get that C/2 intersects its own translate (with a grid point 'a'). Because of symmetry, we get another intersection point. Their halving point is a/2 and is contained in C/2, so 'a' must be contained in C.
Appl 1: The center of planted forests is dark.
Claim: Suppose trees (circles) of width (diameter) 0.16 are planted at every gridpoint in a 13 radius circle. Then origin is not visible from outside.
Proof: Visible ray from origin <=> Strip of width 0.16 contains no gridpoints.
Ex 1: Show that if vol(C)>k2^d, then it contains at least 2k gridpoints.
Appl 2: Approximation by rationals.
Claim: For any 0<t<1 and N natural, exists m, n<=N natural: |t-m/n|<1/nN<1/n^2.
Proof: Take strip thru origin of slope t with heightwidth 2/N: |tx-y|<1/N. Also, let x run from -N-1/2 to N+1/2. Now apply Minkowski.
Ex 2: For any 0<t1, t2<1 and N natural, exists m1, m2, n<=N natural: |ti-mi/n|<1/nsqrt(N).
Def: Lattices - informal!!! Volume is det of matrix formed by basis.
Minkowski thm for lattices: Same but now vol(C)>2^d.vol(L) is required. Proof is same.
Appl 3: Primes of the form 4k+1 are sums of two squares. (Fermat)
Prop: In this case there is a q such that q^2 mod p = -1
Proof: (p-1)!=-1 mod p, because apart from 1 and -1, everybody has a different reciprocal.
But also, we can pair up every x with -1/x, then we get (-1)^(2k)=1, a contradiction.
Proof of appl 3: Take lattice with vectors (1,q) and (0,p). Vol is p. Disc of radius sqrt(2p) has a lattice point because of Minkowski (pi>2). So for i(1,q)+j(0,p) we have i^2+(iq+jp)^2<2p. LHS is 0 mod p, so it must equal p.
Appl 4: Every integer is the sum of four squares. (Lagrange)
Prop: For p prime, there is an a, b such that a^2+b^2 mod p = -1
Proof: There are (p+1)/2 quadratic residues, a^2 and -1-b^2 both have this many values, one is common.
Hw: Take lattice (x,y,z,t): z=ax+by (mod p) and t=bx-ay (mod p). Compute volume and use Minkowski to prove Lagrange thm for primes.
Finally note that: (a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)=
(a_1 b_1 - a_2 b_2 - a_3 b_3 - a_4 b_4)^2 +
(a_1 b_2 + a_2 b_1 + a_3 b_4 - a_4 b_3)^2 +
(a_1 b_3 - a_2 b_4 + a_3 b_1 + a_4 b_2)^2 +
(a_1 b_4 + a_2 b_3 - a_3 b_2 + a_4 b_1)^2.