- Write
*f*(*x*) =*xy*^{3}+*yz*^{3}+*zx*^{3}-*x*^{3}*y*-*y*^{3}*z*-*z*^{3}*x*. First we look for local maxima, so we need to solve ∂*f*/∂*x*= ∂*f*/∂*y*= ∂*f*/∂*z*= 0. Now ∂*f*/∂*x*=*y*^{3}+3*x*^{2}*z*-*z*^{3}-3*x*^{2}*y*. If*y*=*z*, then*f*(*x*,*y*,*z*) = 0 and this is not a maximum. Thus we may divide by*y*-*z*and then ∂*f*/∂*x*= 0 yields*y*^{2}+*yz*+*z*^{2}= 3*x*^{2}. Similarly*x*^{2}+*xz*+*z*^{2}= 3*y*^{2}and*x*^{2}+*xy*+*y*^{2}= 3*z*^{2}. Adding these three equations, we obtain (*x*-*y*)^{2}+ (*y*-*z*)^{2}+ (*z*-*x*)^{2}= 0, which yields*x*=*y*=*z*. This does not give a maximum, because*f*= 0 in this case, and we conclude that the maximum of*f*must occur on the boundary of the region, so at least one of*x*,*y*,*z*is 0 or 1.Let's look at

*f*on the side*x*= 0. Here*f*=*yz*^{3}-*y*^{3}*z*and 0≤*y*≤1, 0≤*z*≤1. To find local maxima, we solve ∂*f*/∂*y*= ∂*f*/∂*z*= 0. This yields*y*=*z*= 0 and*f*= 0, which is not a maximum, so the maximum occurs on the edges of the region considered. If*y*or*z*= 0, we get*f*= 0 which is not a maximum. If*y*= 1, then*f*=*z*^{3}-*z*≤ 0, which won't give a maximum. Finally if*z*= 1, then*f*=*y*-*y*^{3}. Since*df*/*dy*= 1 - 3*y*^{2}, we see that*f*has a maximum at*y*= 1/√3. This gives that the maximum value of*f*on*x*= 0 is 1/√3 -1/√3^{3}= 2√3/9.Similarly if

*y*or*z*= 0, the maximum value of*f*is 2√3/9. Now let's look at*f*on the side*x*= 1. Here*f*=*y*^{3}+*yz*^{3}+*z*-*y*-*y*^{3}*z*-*z*^{3}. Again we first look for local maxima: ∂*f*/∂*y*= 3*y*^{2}+*z*^{3}-1 - 3*y*^{2}*z*. Then ∂*f*/∂*y*= 0 yields either*z*= 1 or 3*y*^{2}=*z*^{2}+*z*+ 1. If*z*= 1, then*f*= 0 which is not a maximum, so 3*y*^{2}=*z*^{2}+*z*+ 1. Similarly 3*z*^{2}=*y*^{2}+*y*+ 1. Adding these two equations, we find that*y*^{2}-*y*/2 +*z*^{2}-*z*/2 = 1. Thus (*y*- 1/2)^{2}+ (*z*- 1/2)^{2}= 3/2. This has no solution in the region considered 0≤*y*≤1, 0≤*z*≤1. Thus*f*must have a maximum on one of the edges. If*y*or*z*is 0, then we are back in the previous case. On the other hand if*y*or*z*is 1, then*f*= 0, which is not a maximum.We conclude that the maximum value of

*f*on 0≤*x*≤1, 0≤*y*≤1, 0≤*z*≤1 is 2√3/9. - For each positive integer
*n*, let*f*(*n*) denote the number of sequences of 1's and 3's that sum to*n*. Then*f*(*n*+ 3) =*f*(*n*+ 2) +*f*(*n*), and we have*f*(1) = 1,*f*(2) = 1, and*f*(3) = 2. Thus*f*(4) =*f*(3) +*f*(1) = 3,*f*(5) =*f*(4) +*f*(2) = 4,*f*(6) = 6, ...,*f*(15) = 189,*f*(16) = 277. Thus the number of sequences required is 277. - Let
*R*denote the specified region, i.e. {(*x*,*y*) |*x*^{4}+*y*^{4}≤*x*^{2}-*x*^{2}*y*^{2}+*y*^{2}}. Then*R*can be described as the region inside the curve*x*^{4}+*x*^{2}*y*^{2}+*y*^{4}=*x*^{2}+*y*^{2}( (*x*,*y*)≠(0, 0)). This can be rewritten as(Now change to polar coordinates: write*x*^{2}+*y*^{2}-*xy*)(*x*^{2}+*y*^{2}+*xy*) =*x*^{2}+*y*^{2}.*x*=*r*cosθ,*y*=*r*sinθ; then the equation becomes (*r*^{2}-*r*^{2}cosθsinθ)(*r*^{2}+*r*^{2}cosθsinθ) =*r*^{2}. Since*r*≠ 0 and 2 cosθsinθ = sin 2θ, we now have*r*^{2}(1 - (1/4)sin^{2}2θ) = 1. Therefore the area*A*of*R*is∬ _{R}*r**drd*θ= ∫ _{0}^{2π}∫_{0}^{(1-(1/4)sin22θ)-1/2}*r**drd*θ = ∫_{0}^{2π}*d*θ/(2(1 - (1/4)sin^{2}2θ))= ∫ _{0}^{π/4}16*d*θ/(3 + cos^{2}2θ) = ∫_{0}^{π/4}16 sec^{2}2θ*d*θ/(4 + 3 tan^{2}2θ).

Now make the substitution 2*z*= √3tan 2θ, so*dz*= √3sec^{2}2θ*d*θ and we obtain*A*= (4/√3)∫_{0}^{∞}*dz*/(1 +*z*^{2}) = 2π/√3. - Ceva's theorem applied to the triangle
*ABC*shows that (*AP*/*PB*)(*BM*/*MC*)(*CN*/*NA*) = 1. Since*BM*=*MC*, we see that*AP*/*PB*=*AN*/*NC*and we deduce that*PN*is parallel to*BC*. Therefore ∠*NPX*= ∠*PCB*= ∠*NAX*and we conclude that*APXN*is a cyclic quadrilateral. Since that opposite angles of a cyclic quadrilateral sum to 180^{o}, we see that ∠*APX*+ ∠*XNA*= 180^{o}, and the result follows. - Let
T = {
*a*_{n}|*n*∈**N**} and for*t*a positive number, let*A*_{t}= {*n*∈**N**|*a*_{n}≥*t*}. Since ∑*a*_{n}= 1 and*a*_{n}≥ 0 for all*n*, we that if δ > 0, then there are only finitely many numbers in T greater than δ, and also*A*_{t}is finite. Thus we may label the nonzero elements of T as*t*_{1},*t*_{2},*t*_{3},..., where*t*_{1}>*t*_{2}>*t*_{3}> ... > 0. We shall use the notation*X*Δ*Y*to indicate the symmetric difference {*X*\*Y*∪*Y*\*X*} of two subsets*X*,*Y*.Consider the sum

∑Note that_{i≥1}(*t*_{i}-*t*_{i+1})|*A*_{ti}Δπ^{-1}*A*_{ti}|.*n*∈*A*_{t}\π^{-1}*A*_{t}if and only if*a*_{n}≥*t*>*a*_{πn}, and*n*∈π^{-1}*A*_{t}\*A*_{t}if and only if*a*_{n}<*t*≤*a*_{πn}. Write*a*_{n}=*t*_{p}and*a*_{πn}=*t*_{q}. We have three cases to examine:-
*a*_{n}=*a*_{πn}. Then*n*does not appear in the above sum. -
*a*_{n}>*a*_{πn}. Then*p*<*q*and*n*is in*A*_{tr}\π^{-1}*A*_{tr}whenever*t*_{p}≥*t*_{r}>*t*_{q}, that is*q*>*r*≥*p*and we get a contribution (*t*_{p}-*t*_{p+1}) + (*t*_{p+1}-*t*_{p+2}) + ... + (*t*_{q-1}-*t*_{q}) =*t*_{p}-*t*_{q}=*a*_{n}-*a*_{πn}= |*a*_{n}-*a*_{πn}|. -
*a*_{n}<*a*_{πn}. Then*p*>*q*and*n*is in π^{-1}*A*_{tr}\*A*_{tr}whenever*t*_{q}≥*t*_{r}>*t*_{p}, that is*p*>*r*≥*q*and we get a contribution (*t*_{q}-*t*_{q+1}) + (*t*_{q+1}-*t*_{q+2}) + ... + (*t*_{p-1}-*t*_{p}) =*t*_{q}-*t*_{p}=*a*_{πn}-*a*_{n}= |*a*_{n}-*a*_{πn}|.

∑because |_{n=1}^{∞}|*a*_{n}-*a*_{πn}| = ∑_{i≥1}(*t*_{i}-*t*_{i+1})|*A*_{ti}Δπ*A*_{ti}|,*A*_{ti}Δπ^{-1}*A*_{ti}| = |*A*_{ti}Δπ*A*_{ti}|. Similarly∑and we deduce that_{n=1}^{∞}|*a*_{n}-*a*_{ρn}| = ∑_{i≥1}(*t*_{i}-*t*_{i+1})|*A*_{ti}Δρ*A*_{ti}|,∑Therefore ∑_{n=1}^{∞}(|*a*_{n}-*a*_{πn}| + ∑_{n=1}^{∞}|*a*_{n}-*a*_{ρn}|) = ∑_{i≥1}(*t*_{i}-*t*_{i+1})(|*A*_{ti}Δπ*A*_{ti}| + |*A*_{ti}Δρ*A*_{ti}|)._{i≥1}(*t*_{i}-*t*_{i+1})(|*A*_{ti}Δπ*A*_{ti}| + |*A*_{ti}Δρ*A*_{ti}|) < ε. We also have ∑_{i≥1}(*t*_{i}-*t*_{i+1})|*A*_{ti}| = 1. Therefore for some*i*, we must have |*A*_{ti}Δπ*A*_{ti}| + |*A*_{ti}Δρ*A*_{ti}| < ε|*A*_{ti}| and the result follows. -
- Multiply
*a*^{4}-3*a*^{2}+ 1 by*b*and subtract (*a*^{3}- 3*a*)(*ab*- 1) to obtain*a*^{3}- 3*a*+*b*. Now multiply by*b*and subtract*a*^{2}(*ab*- 1) to obtain*a*^{2}-3*ab*+*b*^{2}. Thus we want to know when*ab*- 1 divides (*a*-*b*)^{2}- 1, where*a*,*b*are positive integers. We cannot have*a*=*b*, because*a*^{2}- 1 does not divide -1. We now assume that*a*>*b*.Suppose

*ab*- 1 does divide (*a*-*b*)^{2}- 1 where*a*,*b*are positive integers. Write (*a*-*b*)^{2}- 1 =*k*(*ab*- 1), where*k*is an integer. Since (*a*-*b*)^{2}-1≥ 0, we see that*k*is nonnegative. If*k*= 0, then we have (*a*-*b*)^{2}= 1, so*a*-*b*= ±1. In this case,*ab*- 1 does divide*a*^{4}-3*a*^{2}+ 1, because*a*^{4}-3*a*^{2}+1 = (*a*^{2}+*a*- 1)(*a*^{2}-*a*- 1). We now assume that*k*≥1.Now fix

*k*and choose*a*,*b*with*b*as small as possible. Then we have*a*^{2}+*a*(- 2*b*-*kb*) +*b*^{2}+*k*- 1 = 0. Consider the quadratic equation*x*^{2}+*x*(- 2*b*-*kb*) +*b*^{2}+*k*- 1 = 0. This has an integer root*x*=*a*. Let*v*be its other root. Since the sum of the roots is 2*b*+*kb*, we see that*v*is also an integer. Also*av*=*b*^{2}+*k*- 1. Since*b*,*k*≥1, we see that*v*is also positive. We want to show that*v*<*b*; if this was not the case, then we would have*b*^{2}+*k*- 1≥*ab*, that is*k*≥*ab*-*b*^{2}+ 1. We now obtain(This simplifies to*a*-*b*)^{2}-1≥(*ab*-*b*^{2}+ 1)(*ab*- 1).*a*^{2}-*ab*≥(*ab*-*b*^{2}+ 1)*ab*, that is*a*-*b*≥(*ab*-*b*^{2}+ 1)*b*and we obtain (*a*-*b*)(*b*^{2}-1) +*b*≤ 0, which is not the case. Thus*v*<*b*and we have*v*^{2}+*v*(- 2*b*-*k*) +*b*^{2}+*k*- 1 = 0. Set*u*=*b*. Then we have (*u*-*v*)^{2}- 1 =*k*(*uv*- 1), where*u*,*v*are positive and*v*<*b*. By minimality of*b*, we conclude that there are no*a*,*b*such that (*a*-*b*)^{2}- 1 =*k*(*ab*- 1).Putting this altogether, the positive integers required are all

*a*,*b*such that*b*=*a*±1. - Note that for fixed
*x*> 1, the sequence 1/*f*_{n}(*x*) is decreasing with respect to*n*and positive, so the given limit exists which means that*g*is well-defined. Next we show that*g*(*e*^{1/e})≥1/*e*, equivalently lim_{n--> ∞}*f*_{n}(*e*^{1/e})≤*e*. To do this, we show by induction that*f*_{n}(*e*^{1/e})≤*e*for all positive integers*n*. Certainly*f*_{1}(*e*^{1/e}) =*e*^{1/e}≤*e*. Now if*f*_{n}(*e*^{1/e})≤*e*, then*f*_{n+1}(*e*^{1/e}) = (*e*^{1/e})^{fn(e1/e)}≤(*e*^{1/e})^{e}=*e*,*g*(*e*^{1/e})≥1/*e*.We now prove that

*g*(*x*) = 0 for all*x*>*e*^{1/e}; this will show that*g*is discontinuous at*x*=*e*^{1/e}. We need to prove that lim_{n--> ∞}*f*_{n}(*x*) = ∞. If this is not the case, then we may write lim_{n--> ∞}*f*_{n}(*x*) =*y*where*y*is a positive number > 1. We now have*y*= lim_{n--> ∞}*f*_{n}(*x*) = lim_{n--> ∞}*f*_{n+1}(*x*) =*x*^{limn--> ∞fn(x)}=*x*^{y}.*y*=*y*ln*x*and*x*=*y*^{1/y}. Since (*dx*/*dy*)/*x*= (1 - ln*y*)/*y*^{2}, we see by considering the graph of*y*^{1/y}that it reaches its maximum when*y*=*e*, and we deduce that*x*≤*e*^{1/e}. This is a contradiction and we conclude that lim_{n--> ∞}*f*_{n}(*x*) = 0. Thus we have shown that*g*(*x*) is discontinuous at*x*=*e*^{1/e}.

Peter Linnell 2008-11-06