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=2n=∞f (n) -f (n + 2). Since limn→∞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 19m ≡ 1 mod 22014. Therefore n divides 22013, so n = 2k for some positive integer k. Now

192k -1 = (19 - 1)(19 + 1)(192 +1)(194 +1)...(192k-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 ia+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=03ia+k+2b = (1 + i + i2 + i3)ia+2b = 0 and k=03ia+2(b+k) = (1 + i2 + i4 + i6)ia+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) = mr, 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) = xr and n + 1 = yr for some x, y∈ℕ. But then n(n + 2) + 1 = (n + 1)2 = zr where z = y2. Since (n + 1)2 > n(n + 2), we see that z > x and hence zx + 1, because x, z∈ℕ. We deduce that zr≥(x + 1)r > xr + 1, a contradiction and the result follows.

6. (a)
Since A and B are finite subsets of T, we may choose aA and bB so that f (ab) is as large as possible. Suppose we can write gab = cd with cA and dB. Let h = d-1b and db. Note that g, hT. Then hI 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 MS 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 = (xA, yA) and B = (xB, yB). 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 12) where πi : AiBi, and the second term is the negative of the number of pairs 12) where π1 : A1B2 and π2 : A2B1. The configuration of the points implies that every pair of paths 12) where π1 : A1B2 and π2 : A2B1 must intersect. Let 𝓘 ≔{(π12) : π1∩π2≠Ø} (this is the set of all intersecting paths, regardless of their endpoints). Define Φ : 𝓘→𝓘 as follows. If 12)∈𝓘 then Φ((π12)) = (π1',π2') and the new pair of paths is obtained from the old one by switching the tails of π12 after their last intersection point. In particular, the pairs 12) and 1',π2') must appear in different terms of det M. But it is clear that ΦoΦ = id𝓘, 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 {(π12) : 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 = #{(π12) : π1∩π2 = Ø}.)

Peter Linnell 2014-10-29