- Let
*S*denote the sum of the given series. By partial fractions,2(If*n*^{2}-2*n*- 4)/(*n*^{4}+4*n*^{2}+16) = (*n*- 2)/(*n*^{2}-2*n*+ 4) -*n*/(*n*^{2}+ 2*n*+ 4).*f*(*n*) = (*n*- 2)/(*n*^{2}- 2*n*+ 4), then 2*S*= ∑_{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 2*S*=*f*(2) -*f*(4) +*f*(3) -*f*(5) +*f*(4) -*f*(6) + ..., so 2*S*=*f*(2) +*f*(3) and we deduce that*S*= 1/14. - Let
*I*denote the given integral. First we make the substitution*y*=*x*^{2}, so*dy*= 2*xdx*. Then2Now make the substitution*I*= ∫_{0}^{4}(16 -*y*)/(16 -*y*+ √((16 -*y*)(12 +*y*)))*dy*= ∫_{0}^{4}(√16-*y*)/(√(16 -*y*) + √(12 +*y*))*dy*.*z*= 4 -*y*, so*dz*= -*dy*. Then2Adding the last two equations, we obtain 4*I*= ∫_{0}^{4}(√12+*z*)/(√(12 +*z*) + √(16 -*z*))*dz*.*I*= ∫_{0}^{4}*dz*= 4 and hence*I*= 1. - Let
*m*= φ(2^{2014}) = 2^{2013}(here φ(*x*) is Euler's totient function, the number of positive integers <*x*which are prime to*x*). Then 19^{m}≡ 1 mod 2^{2014}. Therefore*n*divides 2^{2013}, so*n*= 2^{k}for some positive integer*k*. Now19we calculate the power of 2 in the above expression. This is 1 + 2 + 1 + 1 + ... + 1 =^{2k}-1 = (19 - 1)(19 + 1)(19^{2}+1)(19^{4}+1)...(19^{2k-1}+ 1);*k*+ 2. Therefore*k*+ 2 = 2014 and it follows that*n*= 2^{2012}. - 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=0}^{3}*i*^{a+k+2b}= (1 +*i*+*i*^{2}+*i*^{3})*i*^{a+2b}= 0 and ∑_{k=0}^{3}*i*^{a+2(b+k)}= (1 +*i*^{2}+*i*^{4}+*i*^{6})*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*i*^{10+20}= -1. On the other hand this sum is also( *i*+*i*^{2}+ ... +*i*^{19})(*i*^{2}+*i*^{4}+ ... +*i*^{38})= *i*(*i*^{19}-1)/(*i*- 1)·(- 1 + 1 - ... - 1)= *i*(-*i*- 1)/(*i*- 1)· - 1 = 1.

This is a contradiction and therefore we have no such tiling. - 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*=*y*^{2}. 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. - (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*^{-1}*b*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*M*^{3}=*I*. Suppose*f*(*M*) >*f*(*I*). Then (*X*=*M*and*Y*=*M*) we obtain either*f*(*M*^{2}) >*f*(*M*) or*f*(*I*) >*f*(*M*), hence*f*(*M*^{2}) >*f*(*M*). Now do the same with*X*=*M*^{2}and*Y*=*M*: we obtain*f*(*M*^{3}) >*f*(*M*^{2}). Since*M*^{3}=*I*, we now have*f*(*I*) >*f*(*M*^{2}) >*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*.

- (a)
- Let
*A*= (*x*_{A},*y*_{A}) and*B*= (*x*_{B},*y*_{B}). Then*d*(*A*,*B*) =( *x*_{B}-*x*_{A}+*y*_{A}-*y*_{B}*x*_{B}-*x*_{A}) - (b)
- By definition
det
*M*=*d*(*A*_{1},*B*_{1})*d*(*A*_{2},*B*_{2}) -*d*(*A*_{1},*B*_{2})*d*(*A*_{2},*B*_{1}). Note that the first term counts all pairs of paths (π_{1},π_{2}) where π_{i}:*A*_{i}→*B*_{i}, and the second term is the negative of the number of pairs (π_{1},π_{2}) where π_{1}:*A*_{1}→*B*_{2}and π_{2}:*A*_{2}→*B*_{1}. The configuration of the points implies that every pair of paths (π_{1},π_{2}) where π_{1}:*A*_{1}→*B*_{2}and π_{2}:*A*_{2}→*B*_{1}must intersect. Let 𝓘 ≔{(π_{1},π_{2}) : π_{1}∩π_{2}≠Ø} (this is the set of all intersecting paths, regardless of their endpoints). Define Φ : 𝓘→𝓘 as follows. If (π_{1},π_{2})∈𝓘 then Φ((π_{1},π_{2})) = (π_{1}',π_{2}') and the new pair of paths is obtained from the old one by switching the tails of π_{1},π_{2}after their*last*intersection point. In particular, the pairs (π_{1},π_{2}) and (π_{1}',π_{2}') must appear in different terms of det*M*. But it is clear that Φ`o`Φ =*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 {(π_{1},π_{2}) : π_{1}∩π_{2}= 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*= #{(π_{1},π_{2}) : π_{1}∩π_{2}= Ø}.)

Peter Linnell 2014-10-29