Lektion 4

Theorem 4.1 Let x, y, z be positive integers such that SGD(x, y, z) = 1. Then the following two statements are equivalent :

(i) x is even and x^{2} + y^{2} = z^{2}, (ii) there exist positive integers a < b such that SGD(a, b) = 1 and

x = 2ab, y = b^{2} − a^{2}, z = b^{2} + a^{2}.

Proof : Left as an exercise.

A triple of relatively prime integers satisfying the equivalent conditions of Theorem 4.1 is called a primitive Pythagorean triple.

Theorem 4.2 Let x, y, z be integers such that x^{4} + y^{4} = z^{4}. Then xyz = 0.

Proof : We consider more generally the equation

x^{4} + y^{4} = z^{2}  (1)

and show that it has no integer solutions such that xyz ≠ 0. The proof is by contradiction and makes use of Fermat’s technique of infinite descent. More precisely, the idea is as follows : we suppose there exists a solution (x, y, z) of (1) for which xyz ≠ 0. Then there must be a solution for which, in addition, SGD(x, y, z) = 1. Given any such ‘primitive’ solution, we then show how one may construct a new one (X, Y, Z) such that |Z| < |z|. But since there must be, among all primitive solutions, one in which |z| is minimised, we thereby obtain the desired contradiction.

So let (x, y, z) be a primitive solution to (1). Then (x^{2}, y^{2}, z) is a primitive Pythagorean triple so, by Theorem 4.1, if we assume WLOG that x is even and y odd, then there exist relatively prime integers a, b such that

x^{2} = 2ab,  (2)
y^{2} = b^{2} − a^{2},  (3)
z = b^{2} + a^{2}.  (4)

We can rewrite (3) as

y^{2} + a^{2} = b^{2},

and so (y, a, b) is also a primitive Pythagorean triple. Since y is odd, so must a be even, and there exist relatively prime integers p, q such that

a = 2pq,  (5)
y = q^{2} − p^{2},  (6)
b = q^{2} + p^{2}.  (7)

Substituting (5) and (7) into (2) yields

x^{2} = 4pq(p^{2} + q^{2}).  (8)

Now p and q are relatively prime, hence one sees easily that both are relatively prime to p^{2} +q^{2}. Thus, the three numbers p, q and p^{2} +q^{2} are pairwise relatively prime. Since their product is, by (8), a perfect square, it follows from the Fundamental Theorem of Arithmetic that each is a perfect square. In other words, there exist pairwise relatively prime integers X, Y, Z such that

p = X^{2}, q = Y^{2}, p^{2} + q^{2} = Z^{2}.

Substituting the first two of these relations into the third yields the relation

X^{4} + Y^{4} = Z^{2},

so we have constructed the desired new primitive solution to (1). It remains to check that |Z| < |z|. But, using (7) and (4), we have

Z^{2} = p^{2} + q^{2} = b < b^{2} + a^{2} = z ≤ z^{2},

as required.

Lecture 5

Theorem 5.1 The only integers x, y such that

y^{2} + 2 = x^{3}  (9)

are x = 3, y = ±5.

‘Proof’ : Write (9) as

(y + \sqrt{−2})(y − \sqrt{−2}) = x^{3}.  (10)

We shall first verify that, for any integer y, the numbers y + \sqrt{−2} and y − \sqrt{−2} have no common factor in

Z[\sqrt{−2}] = \{a + b\sqrt{−2} : a, b ∈ Z\}.

For suppose z := a + b\sqrt{−2} is a common factor. Then z divides

(y + \sqrt{−2}) − (y − \sqrt{−2}) = 2\sqrt{−2},

and, taking squares of absolute values (as complex numbers), we conclude that

a^{2} + 2b^{2} | 8,

as ordinary integers. The only possibilities are thus

(i) a = ±1, b = 0.

(ii) a = ±2, b = 0.

(iii) a = 0, b = ±1.

(iv) a = 0, b = ±2.

In case (i), z = ±1, and hence not a proper factor. In all other cases, there must exist integers c, d such that

y + \sqrt{−2} = (a + b\sqrt{−2})(c + d\sqrt{−2}).

From this it follows, by equating the real and imaginary parts, that

y = ac − 2bd,  (11)
1 = ad + bc.  (12)

Now (12) immediately rules out (ii) and (iv), since in both cases the rhs of (12) is even. But from (9) it follows already that y must be odd (otherwise the lhs of (9) will be even, but not divisible by 4), and then (11) also eliminates (iii).

Hence, we have proven that y ± \sqrt{−2} have no common factor in Z[\sqrt{−2}].

By (10), this implies that each of them must be a cube in Z[\sqrt{−2}].

Thus, there exist integers a, b such that

y + \sqrt{−2} = (a + b\sqrt{−2})^{3}.  (13)

Multiplying out the rhs of (13) and equating the real and imaginary parts yields the two equations

y = a^{3} − 6ab^{2},  (14)
1 = 3a^{2}b − 2b^{3} = b(3a^{2} − 2b^{2}).  (15)

Immediately, (14) implies that

b = 3a^{2} − 2b^{2} = ±1.

Since a, b are integers, the only possibility is b = 1, a = ±1. Substituting these possibilities into (13) gives y = ±5, v.s.v.

OBS !! There is a major gap in the proof, namely the part in italics. What I state there is correct, but it requires a proof. What one would actually like to prove is a generalisation of the Fundamental Theorem of Arithmetic to the ring Z[\sqrt{−2}]. We will return to this issue later in the course. But, for the moment, it is worth remarking that the F.T.A. does not hold in, for example, the ring Z[\sqrt{−5}].

Lektion 6 (?)

Notation : C^{×} denotes the group of non-zero complex numbers under multiplication.

Definition 6.1 : Let G be a finite abelian group. A character of G is a group homomorphism

χ : G → C^{×}.

Note that, since G is finite, χ(g) must be a root of unity for any g ∈ G and any character χ. Indeed, if n · g = 0_{G}, then χ(0) = 1 = χ(ng) = [χ(g)]^{n}, so χ(g) is an n:th root of unity.

Given two characters χ_{1}, χ_{2} of a group G, we can define a third character χ_{1} · χ_{2}, called the ‘product’ of χ_{1} and χ_{2}, as follows :

(χ_{1} · χ_{2})(g) := χ_{1}(g) · χ_{2}(g), ∀ g ∈ G.  (16)

It is easy to check that χ_{1} · χ_{2} is also a character.

Notation : The set of characters of an abelian group G is denoted by Ĝ.

Theorem 6.2 Let G be a finite abelian group. Under the multiplication of characters defined by (16), Ĝ becomes a finite abelian group isomorphic to G.

Proof : The details were given in class. To show that Ĝ is a group (that the multiplication is associative, and that an identity element and inverses exist) is easy. That this group is isomorphic to G is the interesting part. The idea for showing this was as follows : Let

G = C_{1} ⊕ C_{2} ⊕ ⋯ ⊕ C_{k}

be any decomposition of G as a direct sum of cyclic groups. Let n_{i} denote the order of C_{i} and let e_{i} be any group element which generates C_{i}. We think of the set {e_{1}, ..., e_{k}} as a ‘basis’ for G. Then there is a canonical ‘dual basis’ {χ_{1}, ..., χ_{k}} for Ĝ given by

χ_{i}(e_{j}) = \{ \begin{matrix}e^{2πi/n_{i}}, & if i = j, \\ 1, & if i ≠ j. \end{matrix}

Notation/Terminology : The identity element of the character group Ĝ is denoted χ_{0}. It is called the trivial character and is the mapping given by

χ_{0}(g) = 1, ∀ g ∈ G.

The following proposition will prove useful in our study of Dirichlet Lfunctions.

Proposition 6.3 Let G be a finite abelian group, g^{∗} ∈ G and χ^{∗} ∈ Ĝ. Then

(i)

∑_{χ∈Ĝ} χ(g^{∗}) = \{ \begin{matrix}|G|, & if g^{∗} = 0, \\ 0, & if g^{∗} ≠ 0. \end{matrix}

(ii)

∑_{g∈G} χ^{∗}(g) = \{ \begin{matrix}|G|, & if χ^{∗} = χ_{0}, \\ 0, & if χ^{∗} ≠ χ_{0}. \end{matrix}

Proof : (i) was done in class, and (ii) left as an exercise. For (i), the argument is as follows : if g ≠ 0 then there exists at least one character - let’s pick one and denote it χ_{g} - with the property that χ_{g}(g) ≠ 1. This follows from the proof of Theorem 6.2, as outlined above.

Then, since Ĝ is a group, we have that

∑_{χ} χ(g) = ∑_{χ}(χ · χ_{g})(g) = χ_{g}(g) · ∑_{χ} χ(g).

Since χ_{g}(g) ≠ 1, it follows that the sum must be zero, v.s.v.

Lecture 10 (?)

I gave a second proof of Gauss’ reciprocity law, as an alternative to the one in Baker. A handout from the book [1], containing the proof, was given. The idea was to use Gauss sums. The relevant preparatory information on Gauss sums is contained in my old lecture notes, on pages 38-39 (specifically, Proposition 24).

Reference

[1] N. Koblitz, A course in numebr theory and cryptography, Springer 1987 (GTM Series, No. 114).