Introduction to Number Theory (NMAI040), winter term 2017/2018

As I am lecturing in English this term, information here will be in English too. For this course in previous years see here (in Czech, though). Syllabus and annotation in SIS. Here are lectures notes in English, here are some errata. Lectures are on Friday in S4 (3rd floor of the Malostranske square building) in 9-10:30 I built this course on the classical book G. H. Hardy & E. M. Wright: An Introduction to the Theory of Numbers and (almost) everything I am lecturing about can be found there.  In the lectures I may provide further references and literature.
Examination is in oral form (you get a question from the following list, prepare for some time and then explain it to me). Terms of examinations: see SIS (or by appoinment). Questions: will be added.

lecture 1, October 6, 2017. 1. Diophantine approximation. Dirichlet's theorem (DT): part 1 says that |a - p/q| < or = 1/qQ has for any real number a and an integer Q > 1 at least one rational solution p/q with 0 < q < Q, part 2 says that |a - p/q| < q^{-2}has for any irrational real number a infinitely many rational solutions p/q. I mentioned (without proofs) three other results of P. Dirichlet in NT: i) theorem on prime numbers in arithmetic progression (which is his most famous and most important), ii) asymptotic of the sum sum_{n<x}tau(n) where tau(n) counts divisors of n, iii) structural theorem on the group of units in the ring of integers of a number field (Dirichlet's theorem on units, I will state it precisely later when we discuss solutions of the Pell equation). As a corollary of part 1 of DT we proved the Euler-Fermat theorem: every prime number of the form p = 1+4n is a sum of two squares. We used a Lemma: -1 is quadratic residue modulo such p. Proof: if a = ((p - 1)/2)! then a^2 is -1 modulo p. 

lecture 2, October 13, 2017. A corollary of part 2 of DT: every Pell equation x^2 - dy^2 = 1 (d > 0 and is not a square) has a nontrivial solution x, y in Z;  proof. The Farey fractions - another way how to prove part 2 of Dirichlet's theorem, in fact even that |a - p/q| < 1/2q^2 has for any irrational real number a infinitely many rational solutions p/q (an exercise for you). The best constant in the denominator is 5^{1/2}, as Hurwitz proved; so |a - p/q| < 1/5^{1/2}q^2 has for any irrational real number a infinitely many rational solutions p/q but for a = (5^{1/2} - 1)/2 and any A > 5^{1/2},  |a - p/q| < 1/Aq^2 has only finitely many rational solutions p/q (without proof). Algebraic and transcendental (real) numbers. Existence of transcendental numbers via Liouville's inequality: |a - p/q| > c/q^n for any real algebraic number a of deg(a) = n > 1 and any fraction p/q (here c = c(a) > 0 is a constant); proof. Thus the number beta := sum 1/10^{n!} is transcendental; proof as an exercise. I mentioned the theorem of Roth: the n in Liouville's inequality may be replaced with 2 + epsilon.

lecture 3, October 20, 2017. I mentioned Thue's inequality from 1908: |a - p/q| >>_{a, ep}q^{-n/2 - 1 - ep} for any real algebraic number a with deg(a) = n > 1, any ep > 0 and any fraction p/q. It implies that every Thue's equation F(x, y) = c has only finitely many integral solutions x, y (here c is an integer and F is a homogeneous integral polynomial that is irreducible in Q[x, y] and has degree at least 3, e.g. x^3 - 2y^3). Hermite's theorem with Hilbert's proof: the number e = 2.71828... is transcendental. 2. Diophantine equations.  Theorem (Putnam, Davis, Robinson, Matiyasevich): solvability of a given Diophantine equation P = 0 is (algorithmically) undecidable (here P is in Z[x_1, ..., x_n] and we seek a solution in Z), without proof of course. Exercise: show that one can (algorithmically) reduce the solvability problem to polynomials with degree at most 4. Pythagorean triples: integral solutions of x^2 + y^2 = z^2. Theorem: (x, y, z) is a primitive P. triple (x, y, z coprime, positive, x even y, z odd) <=> x = 2ab, y = a^2 - b^2, z = a^2 + b^2 with a > b > 0 being coprime integers, one odd and the other even, proof next time.

lecture 4, October 27, 2017. I indicated by an example the reduction from the exercise, a smaller example is here: x^5 + y^6 = 22334455667788 is solvable <=> (a - x^2)^2 + (b - a^2)^2 + (c - bx)^2 + (d - y^2)^2 + (e - d^2)^2 + (f - de)^2 + (c + f - 22334455667788)^2 = 0 is solvable (Mr. Skotnica pointed out that I had omitted the last square). Proof of the theorem about primitive P. triples. Fermat's theorem (FLT for n = 4): x^4 + y^4 = z^2 has no nontrivial solution (solution in N), proof by infinite descent. The Stothers-Mason theorem (SMT): if a, b, c in C[t] are coprime polynomials, not all of them constant, such that a + b = c then max(deg a, deg b, deg c) < r(abc) where r(p) is the number of distinct roots of a polynomial p, proof next time. I concluded with showing that the SMT => if a, b, c in C[t] are coprime polynomials, not all of them constant, and n > 0 is an integer  such that a^n + b^n = c^n then n < 3. This is FLT (Fermat's last theorem) for polynomials. But I promised to show next time for it a more direct proof, not using the SMT.

lecture 5, November 3, 2017. Proof of the SMT.  A direct proof of the FLT for polynomials: if a, b, c in C[t] are coprime polynomials, not all of them constant, and n > 0 is an integer  such that a^n + b^n = c^n then n < 3. Proof goes as follows. Suppose for contradiction that n > 2 and that max(deg a, deg b) is minimum. For  odd n we have factorization c^n = a^n + b^n = (a + b)(a + bd)(a + bd^2) ... (a + bd^{n-1}) where d is a primitive n-th root of 1 (for even n we switch to c^n = a^n - b^n). Since the factors are coprime and C[t] is a UFD (and every unit in C[t] is an n-th power), a + bd^j = e_j^n for every j = 0, 1, ..., n-1 and some polynomials e_j. This and the identity (a + bd^2) + d(a + b) = (1 + d)(a + bd) give a smaller solution - a contradiction. Examples of large smallest nontrivial solutions of Pell equations. Theorem: S = {positive solutions a + bd^{1/2} of the Pell equation x^2 - dy^2 = 1}(with the usual multiplication of real numbers) is the infinite cyclic group and R = {all solutions} = S times Z_2. Proof next time.

lecture 6, November 10, 2017.

November (listopad = the month when leaves (listy) fall (padají)) 2017