By Andrew Baker

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Hence if g ∈ G, gx = g(ky) = (g ∗ k)y ∈ OrbG (y) and so OrbG (x) ⊆ OrbG (y). By (b), x ∈ OrbG (y) and so we also have OrbG (y) ⊆ OrbG (x). This gives OrbG (y) = OrbG (x). Conversely, if OrbG (y) = OrbG (x) then y ∈ OrbG (y) = OrbG (x). 32. Let X = be the square with vertices A, B, C, D and let G = Sym( ). Determine StabG (x) and OrbG (x) where (a) x is the vertex A; (b) x is the midpoint M of AB; (c) x is the point P on AB where AP : P B = 1 : 3. Solution. 16. We will write permutations of the vertices in cycle notation.

B) If there is a surjection m −→ n then m n. (c) If there is a bijection m −→ n then m = n. Proof. (a) We will prove this by Induction on n. Consider the statement P (n) : For m ∈ N0 , if there is a injection m −→ n then m n. When n = 0, there is exactly one function ∅ −→ ∅ (the identity function) and this is a bijection; if m > 0 then there are no functions m −→ ∅. So P (0) is true. Suppose that P (k) is true for some k ∈ N0 and let f : m −→ k + 1 be an injection. We have two cases to consider: (i) k + 1 ∈ im f , (ii) k + 1 ∈ / im f .

A) Prove the identities n+ n2 + 1 = 2n + ( n2 + 1 − n) = 2n + 1 √ . n + n2 + 1 √ √ (b) Show that [ n2 + 1] = n and that the infinite continued fraction expansion of n2 + 1 is [n; 2n]. √ √ (c) Show that [ n2 + 2] = n and that the infinite continued fraction expansion of n2 + 2 is [n; n, 2n]. PROBLEM SET 1 27 √ (d) √ Show that [ n2 + 2n] = n and that the infinite continued fraction expansion of n2 + 2n is [n; 1, 2n]. 1-23. Find the fundamental solutions of Pell’s equation x2 − dy 2 = 1 for each of the values d = 5, 6, 8, 11, 12, 13, 31, 83.