- Let
*I*denote the value of the integral. We make the substitution*y*= π/2 -*x*. Then*dx*= -*dy*, and as*x*goes from 0 to π/2,*y*goes from π/2 to 0. Also sin(π/2 -*x*) = cos*x*and cos(π/2 -*x*) = sin*x*. Thus*I*= ∫_{0}^{π/2}(sin^{4}*x*+ sin*x*cos^{3}*x*+ sin^{2}*x*cos^{2}*x*+ sin^{3}*x*cos*x*)/(sin^{4}*x*+ cos^{4}*x*+ 2 sin*x*cos^{3}*x*+ 2 sin^{2}*x*cos^{2}*x*+ 2 sin^{3}*x*cos*x*)*dx*.*I*= ∫_{0}^{π/2}*dx*. Therefore*I*= π/4. - We necessarily have
*x*≥ - 2. Also the left hand side becomes negative for*x*≥2. Therefore we may assume that*x*= 2 cos*t*for 0≤*t*≤π. After making this substitution, the equation becomes cos 3*t*+ cos(*t*/2) = 0. Using a standard trig formula ( 2 cos*A*cos*B*= cos(*A*+*B*) + cos(*A*-*B*)), this becomes cos(7*t*/4)cos(5*t*/4) = 0. This results in the solutions*t*= 2π/5, 2π/7, 6π/7. Therefore*x*= 2 cos(2π/5), 2 cos(2π/7), 2 cos(6π/7). - We make
*a*,*b*,*c*,*d*,*e*be the roots of the quintic equation*x*^{5}+*px*^{4}+*qx*^{3}+*rx*^{2}+*sx*+*t*= 0. Using the first and last equations, we get*p*=*t*= 1. Let*Z*= {*a*,*b*,*c*,*d*,*e*}. Then2so*q*= ∑_{u, v∈Z, u≠v}*uv*= (*a*+ ... +*e*)^{2}- (*a*^{2}+ ... +*e*^{2}) = -14,*q*= -7. Next*s*=*abcde*(1/*a*+ ... + 1/*e*) = -1/ - 1 = 1. Finally*r*=*abcde*(∑_{u, v∈Z, u≠v}*uv*)= *abcde*((1/*a*+ ... +1/*e*)^{2}- (1/*a*^{2}+ ... +1/*e*^{2}))= -14,

so*r*= -7.Similarly

*s*= 1 and*r*= -7. Therefore*a*,*b*,*c*,*d*,*e*are the roots of*x*^{5}+*x*^{4}-7*x*^{3}-7*x*^{2}+*x*+ 1 = 0. By inspection, -1 is a root and the equation factors as(Using the quadratic formula, it follows that*x*+ 1)(*x*^{4}-7*x*^{2}+1) = (*x*+ 1)(*x*^{2}-3*x*+ 1)(*x*^{2}+ 3*x*+ 1).*a*,*b*,*c*,*d*,*e*are (in whatever order you like)-1,(±3±√5)/2. - We repeatedly use the fact that if
*n*is a positive integer and*a*∈ℤ is prime to*n*, then*a*^{φ(n)}≡1 mod*n*where φ is Euler's totient function.We first show that

*f*(*n*)≡3 mod 4 for all*n*≥1. We certainly have*f*(1)≡3 mod 4. Since*f*(*n*) is always odd, we see that*f*(*n*+ 1)≡3^{f(n)}≡3^{f(n-1)}≡*f*(*n*) mod 4 and we deduce that*f*(*n*)≡3 mod 4 for all*n*≥1.Now we show that

*f*(*n*)≡*f*(3) mod 25 for all*n*≥3. First observe that*f*(*n*+ 1)≡3^{f(n)}≡3^{f(n-1)}≡*f*(*n*) mod 5 for*n*≥2, provided*f*(*n*) =*f*(*n*- 1) mod 4, which is true by the previous paragraph. It follows that*f*(*n*+ 1)≡*f*(*n*) mod 20 for all*n*≥2. Therefore*f*(*n*+ 1)≡3^{f(n)}≡3^{f(n-1)}≡*f*(*n*) mod 25, provided*n*≥3, and our assertion is proven. Since the last two digits of*f*(3) are 87, the last two digits of*f*(2012) are also 87. - Let
*f*(*n*) = 1/(ln*n*) - (1/ln*n*)^{(n+1)/n}. Then*f*(*n*)ln*n*= 1 - (ln*n*)^{-1/n}. Assume that*n*> 27. Since ln*n*>*e*, we see that*f*(*n*)ln*n*> 1 -*e*^{-1/n}. Therefore*nf*(*n*)ln*n*>*n*(1 -*e*^{-1/n}). By L'hôpital's rule, lim_{n→∞}*n*(1 -*e*^{-1/n}) = 1. Therefore*nf*(*n*)ln*n*> 1/2 for sufficiently large*n*, so*f*(*n*) > 1/(2*n*ln*n*). Since ∑_{n=2}^{∞}1/(*n*ln*n*) is divergent, it follows that ∑_{n=2}^{∞}*f*(*n*) is also divergent. - We shall prove by induction that
*a*_{n}=*p*if*p*is a prime and*n*=*p*^{m}for some positive integer*m*, and 1 otherwise. This is clear in the case*n*=*p*^{m}, because then there are exactly*m*- 1 nontrivial divisors of*p*^{m}, and each contributes*p*to the denominator of the displayed fraction. The case*n*=*pq*, where*p*,*q*are distinct primes, is also clear, because then*p*and*q*are the only nontrivial divisors of*n*, and they contribute*p*and*q*respectively to the denominator.Now assume that

*n*is neither a prime power, nor a product of two distinct primes, and assume the result is true for all smaller values of*n*. Then we may write*n*=*pm*, where*p*is a prime and*m*is not a prime power. Write*m*=*p*^{a}*k*, where*a*is a nonnegative integer and*k*is prime to*p*. If*d*|*n*, then either*d*|*m*or*d*=*p*^{a+1}*r*, where*r*|*k*. Note that in the latter case,*d*is a prime power only when*r*= 1. Therefore*a*_{n}=*a*_{m}*p*/*a*_{pa+1}= 1*p*/*p*= 1.*a*_{999000}= 1. - Let 0 and
*I*denote the zero and identity 2×2 matrices respectively. Let*A*denote one of the three matrices. The result is clear if*A*= 0 or*I*, because every matrix commutes with 0 and*I*. Next note that*A*^{i}=*A*^{j}, where 0 <*i*<*j*< 5, and we see that the minimum polynomial of*A*divides*x*^{j}-*x*^{i}.Suppose 0 is not an eigenvalue of

*A*. Then*A*is invertible and it follows that*A*^{j-i}=*I*, in particular*I*is one of the matrices. Since*I*commutes with all matrices, the result follows in this case.Thus we may assume that 0 is an eigenvalue of

*A*. Next suppose both the eigenvalues of*A*are 0 (i.e.*A*has a repeated eigenvalue 0). Then the Jordan canonical form of*A*is either 0 or( 0 1 0 0 ) In both case,

*A*^{2}= 0, and since 0 commutes with all matrices, the result follows in this case.Therefore we may assume that

*A*has one eigenvalue 0 and another eigenvalue λ≠ 0. Since the minimum polynomial of*A*divides*x*^{j}-*x*^{i}where 0 <*i*<*j*< 5, we see that the possibilities for λ are 1, -1, or ω where ω is a primitive cube root of 1. Since the eigenvalues of*A*are distinct, it is diagonalizable and in particular, its Jordan canonical form will be( λ 0 0 0 ) If λ = ω, then {

*A*,*A*^{2},*A*^{3}} are three distinct commuting matrices, and the result is proven in this case. Thus we may assume that the Jordan canonical form for*A*is( ± 1 0 0 0 ) Now not all the

*A*_{i}can have Jordan canonical form( 1 0 0 0 ) because then tr(

*A*_{1}+*A*_{2}+*A*_{3}) = 3, so at least one of the matrices, say*A*_{1}, has trace -1. It should be pointed out that we may assume that the*A*_{i}are distinct, if not, then the three matrices come from {*A*_{1},*A*_{1}^{2}}, and since*A*_{1}commutes with*A*_{1}^{2}, the result follows in this case.Suppose tr(

*A*_{2}) = -1 and tr(*A*_{3}) = 1. Then*A*_{1}^{2}=*A*_{2}^{2}=*A*_{3}and*A*_{3}commutes with*A*_{1}and*A*_{2}, and the result is proven in this case.Finally suppose tr(

*A*_{2}) = tr(*A*_{3}) = 1. Then without loss of generality, we may assume that*A*_{1}^{2}=*A*_{2}, and so*A*_{1}= -*A*_{2}. Thus -*A*_{3}≠*A*_{1}or*A*_{2}. Since*A*_{2}*A*_{3}= -*A*_{1}*A*_{3}, we see that*A*_{2}*A*_{3}=*A*_{1}or*A*_{2}. Similarly*A*_{3}*A*_{2}=*A*_{1}or*A*_{2}. Since tr(*A*_{2}*A*_{3}) = tr(*A*_{3}*A*_{2}), we deduce that*A*_{2}*A*_{3}=*A*_{3}*A*_{2}. Also*A*_{1}*A*_{2}=*A*_{2}*A*_{1}, and the result follows.

Peter Linnell 2012-11-03