- Let
*A*= {(*x*,*y*) | 0≤*x*≤2, 0≤*y*≤3, 3*x*≤2*y*} and*B*= {(*x*,*y*) | 0≤*x*≤2, 0≤*y*≤3, 3*x*≤2*y*}. Let*I*= ∫_{0}^{3}∫_{0}^{2}1/(1 + max(3*x*, 2*y*))^{2}*dxdy*. Then max(3*x*, 2*y*) = 2*y*for (*x*,*y*)∈*A*and max(3*x*, 2*y*) = 3*x*for*x*∈*B*.

Therefore

*I*= ∬ _{A}1/(1 + 2*y*)^{2}*dA*+ ∬_{B}1/(1 + 3*x*)^{2}*dA*= ∫ _{0}^{3}∫_{0}^{2y/3}1/(1 + 2*y*)^{2}*dxdy*+ ∫_{0}^{2}∫_{0}^{3x/2}1/(1 + 3*x*)^{2}*dydx*= ∫ _{0}^{3}2*y*/(3(1 + 2*y*)^{2})*dy*+ ∫_{0}^{2}3*x*/(2(1 + 3*x*)^{2})*dx*= ∫ _{0}^{3}1/(3(1 + 2*y*)) - 1/(3(1 + 2*y*)^{2})*dy*+ ∫_{0}^{2}1/(2(1 + 3*x*)) - 1/(2(1 + 3*x*)^{2})*dx*= [(ln(1 + 2 *y*))/6 + 1/(6(1 + 2*y*))]_{0}^{3}+ [(ln(1 + 3*x*))/6 + 1/(6(1 + 3*x*))]_{0}^{2}= (ln 7)/6 + 1/42 - 1/6 + (ln 7)/6 + 1/42 - 1/6 - (7 ln 7 - 6)/21.

- Let
*A*=( 4 -3 2 -1 ) We want to calculate powers of

*A*, and to do this it is useful to find the Jordan canonical form of*A*. The characteristic polynomial of*A*is det(*xI*-*A*) = (*x*- 4)(*x*+ 1) + 6 =*x*^{2}- 3*x*+ 2 which has roots 1,2. Set**u**=( 1 1 ) An eigenvector corresponding to 1 is

**u**and an eigenvector corresponding to 2 is( 3 2 ) Set

*P*=( 1 3 1 2 ) and

*D*= diag(1, 2) (diagonal matrix with 1,2 on the main diagonal). Then*P*^{-1}=( -2 3 1 -1 ) and

*P*^{-1}*AP*=*D*. Let**v**=( 1 0 ) and let

^{T}denote transpose. Since*A*=*PDP*^{-1}, we find that(θ ^{100}**v**)^{T}= *A*^{100}**v**+ (*A*^{99}+*A*^{98}+ ... +*A*+*A*^{0})**u**= *PD*^{100}*P*^{-1}**v**+*P*(*D*^{99}+*D*^{98}+ ... +*D*+*D*^{0})*P*^{-1}**u**

= *P*( 1 0 0 2 ^{100}) *P*^{-1}**v**+ *P*( 100 0 0 2 ^{100}- 1) *P*^{-1}**u**= ( 98 + 3·2 ^{100}98 + 2·2 ^{100}) Thus θ

^{100}(1, 0) = (98 + 3·2^{100}, 98 + 2·2^{100}). - Let
*g*(*x*) denote the power series in*x*1 - (Then for 2≤*x*+*x*^{2}+ ... +*x*^{n}) + (*x*+*x*^{2}+ ... +*x*^{n})^{2}- ... + (- 1)^{n}(*x*+*x*^{2}+ ... +*x*^{n})^{n}+ ....*r*≤*n*, the coefficient of*x*^{r}in*f*(*x*) is the same as the coefficient of*x*^{r}in*g*(*x*). Since*x*+*x*^{2}+ ... +*x*^{n}=*x*(1 -*x*^{n})/(1 -*x*), we see that*g*(*x*) is a geometric series with ratio between successive terms -*x*(1 -*x*^{n})/(1 -*x*), hence its sum is1/(1 +clearly the coefficient of*x*(1 -*x*^{n})/(1 -*x*)) = (1 -*x*)/(1 -*x*^{n+1}) = (1 -*x*)(1 +*x*^{n+1}+*x*^{2n+2}+ ...).*x*^{r}in the above is 0 for 2≤*r*≤*n*, which proves the result. - Write
[τ
*n*] =*p*. Then*p*is the unique integer satisfying*p*< τ*n*<*p*+ 1 because*p*≠τ*n*(otherwise τ =*p*/*n*, a rational number), that is*p*/τ <*n*<*p*/τ + 1. Since 1/τ = τ - 1, we see that*p*τ -*p*<*n*<*p*τ -*p*+ 1 and we deduce that*n*+*p*- 1 <*p*τ <*n*+*p*. Therefore [*p*τ] =*n*+*p*- 1 and hence [τ[τ*n*] + 1] =*n*+*p*. But τ^{2}*n*= τ*n*+*n*, consequently [τ^{2}*n*] =*p*+*n*and the result follows. - Suppose
*x*∈**R**and θ(*x*)≤ - 1. Fix*y*∈**R**with*y*<*x*. Then if*n*is a positive integer and*x*>*p*_{1}> ... >*p*_{n}>*y*, we have for 1≤*i*≤*n*θ( *x*)≥θ( *x*)^{3}> θ(*p*_{1}),θ( *p*_{i})> θ( *p*_{i})^{3}> θ(*p*_{i+1}),θ( *p*_{n})> θ( *p*_{n})^{3}> θ(*y*),

and we deduce that θ(*x*)θ(*p*_{1})^{2n-2}> θ(*y*), for all*n*. this is not possible, so θ(*x*) > - 1 for all*x*∈**R**. The same argument works if 0≤θ(*y*) < θ(*x*)≤1. - We will concentrate on the bottom left hand corner of the square and
determine the area
*A*of that portion of the square that can be painted by the brush, and then multiply that by 4. We make the bottom of the square the*x*-axis and the left hand side of the square the*y*-axis. The equation of a line of length 4 from (*a*, 0) to the*y*-axis is*x*/*a*+*y*/√(16 -*a*^{2}) = 1, that is*y*= (1 -*x*/*a*)√(16 -*a*^{2}). For fixed*x*, we want to know the maximum value*y*can take by varying*a*. To do this, we differentiate*y*with respect to*a*and then set the resulting expression to 0. Thus we need to solve(On multiplying by √(16 -*x*/*a*^{2})√(16 -*a*^{2}) -*a*(1 -*x*/*a*)/√(16 -*a*^{2}) = 0.*a*^{2}) and simplifying, we obtain 16*x*=*a*^{3}and hence*dx*/*da*= 3*a*^{2}/16. Therefore*A*= ∫ _{x=0}^{x=4}(1 -*x*/*a*)√(16 -*a*^{2})*dx*= ∫_{a=0}^{a=4}(1 -*x*/*a*)√(16 -*a*^{2})*dx*/*da**da*= ∫ _{a=0}^{a=4}3*a*^{2}(1 -*a*^{2}/16)√(16 -*a*^{2})/16*da*= ∫_{0}^{4}3*a*^{2}(16 -*a*^{2})^{3/2}/256*da*.

This ia a standard integral which can be evaluated by a trigonometric substitution. Specifically we set*a*= 4 sin*t*, so*da*/*dt*= 4 cos*t*and we find that*A*= ∫ _{0}^{π/2}48 cos^{4}*t*sin^{2}*t**dt*= ∫_{0}^{π/2}6 sin^{2}2*t*(1 + cos 2*t*)*dt*= ∫ _{0}^{π/2}3(1 - cos 4*t*)*dt*= 3π/2.

We conclude that the total area that can be painted by the brush is 6πin^{2}. - Note that if
*p*is a prime, then*f*(*p*) =*p*. Thus*f*(100) =*f*(2^{2}·5^{2}) = 4 + 10 = 14,*f*(2·7) = 2 + 7 = 9,*f*(3^{2}) = 3·2 = 6. Therefore*g*(100) = 6. Next*f*(10^{10}) =*f*(2^{10}·5^{10}) = 20 + 50 = 70,*f*(2·5·7) = 14,*f*(2·7) = 2 + 7 = 9,*f*(3^{2}) = 3·2 = 6. Therefore*g*(10^{10}) = 6.Since

*f*(*p*) =*p*if*p*is prime, we see that*g*(*p*) =*p*also and thus primes cannot have property H. Note that if*r*,*s*are coprime, then*g*(*rs*)≤*f*(*r*)*s*. Suppose*n*has property H and let*p*be a prime such that*p*^{2}divides*n*, so*n*=*p*^{k}*r*where*k*≥2 and*r*is prime to*p*. It is easy to check that if*p*^{k}> 9, then*p*^{k}> 2*pk*, that is*f*(*p*^{k}) <*p*^{k}/2, thus*f*(*n*) <*n*/2 and we see that*n*cannot have property H. Also if*p*,*q*are distinct odd primes and*pq*> 15, then*f*(*pq*) <*pq*/2 and so if*n*=*pqr*with*r*prime to*pq*, then we see again that*n*cannot have property H.The only cases to be considered now are

*n*= 9, 15, 45. By direct calculation, 9 has property H, but 15 and 45 do not. So the only positive odd integer larger than 1 that has property H is 9.

Peter Linnell 2007-06-05