1. Let S denote the sum of the given series. By partial fractions,
   2(n² -2n - 4)/(n⁴ +4n² +16) = (n - 2)/(n² -2n + 4) -n/(n² + 2n + 4).
   If f (n) = (n - 2)/(n² - 2n + 4), then 2S = ∑_n=2^n=∞f (n) -f (n + 2). Since lim_n→∞f (n) = 0, it follows by telescoping series that the series is convergent and 2S = f (2) -f (4) + f (3) -f (5) + f (4) -f (6) + ..., so 2S = f (2) + f (3) and we deduce that S = 1/14.

2. Let I denote the given integral. First we make the substitution y = x², so dy = 2xdx. Then
   2I = ∫₀⁴(16 -y)/(16 -y + √((16 -y)(12 + y))) dy = ∫₀⁴(√16-y)/(√(16 -y) + √(12 + y)) dy.
   Now make the substitution z = 4 -y, so dz = -dy. Then
   2I = ∫₀⁴(√12+z)/(√(12 + z) + √(16 -z)) dz.
   Adding the last two equations, we obtain 4I = ∫₀⁴dz = 4 and hence I = 1.

3. Let m = φ(2²⁰¹⁴) = 2²⁰¹³ (here φ(x) is Euler's totient function, the number of positive integers < x which are prime to x). Then 19^m ≡ 1 mod 2²⁰¹⁴. Therefore n divides 2²⁰¹³, so n = 2^k for some positive integer k. Now
   19^2k -1 = (19 - 1)(19 + 1)(19² +1)(19⁴ +1)...(19^2k-1 + 1);
   we calculate the power of 2 in the above expression. This is 1 + 2 + 1 + 1 + ... + 1 = k + 2. Therefore k + 2 = 2014 and it follows that n = 2²⁰¹².

4. Put i^a+2b in the square in the (a, b) position. Note that the sum of all the entries in a 4×1 or 1×4 rectangle is zero, because ∑_k=0³i^a+k+2b = (1 + i + i² + i³)i^a+2b = 0 and ∑_k=0³i^a+2(b+k) = (1 + i² + i⁴ + i⁶)i^a+2b = 0. Therefore if we have a tiling with 4×1 and 1×4 rectangles, the sum of the entries in all 361 squares is the value on the central square, namely i¹⁰⁺²⁰ = -1. On the other hand this sum is also

   (i + i² + ... + i¹⁹)(i² + i⁴ + ... + i³⁸) = i(i¹⁹ -1)/(i - 1)·(- 1 + 1 - ... - 1) = i(-i - 1)/(i - 1)· - 1 = 1.

   
   
   This is a contradiction and therefore we have no such tiling.

5. Suppose by way of contradiction we can write n(n + 1)(n + 2) = m^r, where n∈ℕ and r≥2. If a prime p divides n(n + 2) and n + 1, then it would have to divide n + 1, and n or n + 2, which is not possible. Therefore we may write n(n + 2) = x^r and n + 1 = y^r for some x, y∈ℕ. But then n(n + 2) + 1 = (n + 1)² = z^r where z = y². Since (n + 1)² > n(n + 2), we see that z > x and hence z≥x + 1, because x, z∈ℕ. We deduce that z^r≥(x + 1)^r > x^r + 1, a contradiction and the result follows.

6. (a)

   Since A and B are finite subsets of T, we may choose a∈A and b∈B so that f (ab) is as large as possible. Suppose we can write g ≔ab = cd with c∈A and d∈B. Let h = d^-1b and d≠b. Note that g, h∈T. Then h≠I and we see that either f (gh^-1) > f (g) or f (gh) > f (g). This contradicts the maximality of f (ab). Therefore d = b and because b is an invertible matrix, we deduce that a = c and the result is proven.

   (b)

   Set

   M =

   (

   	-1	-1
   	1	0

   )

   Then M∈S and M³ = I. Suppose f (M) > f (I). Then (X = M and Y = M) we obtain either f (M²) > f (M) or f (I) > f (M), hence f (M²) > f (M). Now do the same with X = M² and Y = M: we obtain f (M³) > f (M²). Since M³ = I, we now have f (I) > f (M²) > f (M) > f (I), a contradiction. The argument is similar if we start out with f (M) < f (I). This shows that there is no such f.

7. (a)

   Let A = (x_A, y_A) and B = (x_B, y_B). Then

   d (A, B) =

   (

   	xB -xA + yA -yB
   	xB -xA

   )

   (b)

   By definition det M = d (A₁, B₁)d (A₂, B₂) -d (A₁, B₂)d (A₂, B₁). Note that the first term counts all pairs of paths (π₁,π₂) where π_i : A_i→B_i, and the second term is the negative of the number of pairs (π₁,π₂) where π₁ : A₁→B₂ and π₂ : A₂→B₁. The configuration of the points implies that every pair of paths (π₁,π₂) where π₁ : A₁→B₂ and π₂ : A₂→B₁ must intersect. Let 𝓘 ≔{(π₁,π₂) : π₁∩π₂≠Ø} (this is the set of all intersecting paths, regardless of their endpoints). Define Φ : 𝓘→𝓘 as follows. If (π₁,π₂)∈𝓘 then Φ((π₁,π₂)) = (π₁',π₂') and the new pair of paths is obtained from the old one by switching the tails of π₁,π₂ after their last intersection point. In particular, the pairs (π₁,π₂) and (π₁',π₂') must appear in different terms of det M. But it is clear that ΦoΦ = id_𝓘, therefore Φ is an involution. This implies that all intersecting pairs of paths must cancel each other, and that the only pairs which contribute to the determinant are those from the set {(π₁,π₂) : π₁∩π₂ = Ø}. Since all the latter pairs can appear only with positive sign (in the first term of det M), this finishes the solution. (In fact, we proved that det M = #{(π₁,π₂) : π₁∩π₂ = Ø}.)