- Let
*x*and*y*meters denote the distance between the walker and the jogger when the dog returns to the walker for the*n*th and*n*+ 1st times respectively. Then starting at the walker for the*n*th time, it will take the dog*x*seconds to reach the jogger, whence the jogger will be distance 2*x*from the walker, and then it will take the dog a further 2*x*/4 seconds to return to the walker. Thus*y*=*x*+*x*+ 2*x*/4 = 5*x*/2, and also the time taken for the dog to return to the walker will be*x*+ 2*x*/4 = 3*x*/2 seconds. It follows that the total distance travelled by the dog to return to the walker for the*n*th time is3*(3/2)(1 + 5/2 + (5/2)meters. Therefore^{2}+ ... + (5/2)^{n-1}) = 9((5/2)^{n}- 1)/(2(5/2 - 1))*f*(*n*) = 3((5/2)^{n}- 1). - By examining the numbers 5,10,15,20,25,30,35,40 (at most 40 and
divisible by 5), we see that the power of 5 dividing 40! is 9. Also
it is easy to see that the power of 2 dividing 40! is at least 9.
Therefore 10
^{9}divides 40! and we deduce that*p*=*q*=*r*=*s*=*t*=*u*=*v*=*w*=*x*= 0.Next note that 999999 divides 40!. This is because 999999 = 9*111111 = 9*111*1001 = 9*3*37*11*91 = 27*7*11*13*37. Since 999999 = 10

^{6}- 1, it follows that if we group the digits of 40! in sets of six starting from the units digit and working right to left, then the sum of these digits will be divisible by 999999. Therefore*abcdef*+ 283247 + 897734 + 345611 + 269596 + 115894 + 272000 is divisible by 999999, that is*abcdef*+ 2184082 = 999999*y**y*. Clearly*y*= 3 and we conclude that*abcdef*= 815915. - By symmetry,
*f*(*x*) is twice the integral over the triangle bounded by*u*=*v*,*v*= 0 and*u*=*x*, so*f*(*x*) = 2∫_{0}^{x}∫_{0}^{u}*e*^{u2v2}*dvdu*and it follows that*f'*(*x*) = 2∫_{0}^{x}*e*^{x2v2}*dv*. Set*t*=*xv*, so*dv*=*dt*/*x*and*f'*(*x*) = (2/*x*)∫_{0}^{x2}*e*^{t2}*dt*. Therefore*f''*(*x*) = (-2/*x*^{2})∫_{0}^{x2}*e*^{t2}*dt*+ 2*x*(2/*x*)*e*^{x4}.*xf''*(*x*) +*f'*(*x*) = 4*xe*^{x4}and hence 2*f''*(2) +*f'*(2) = 8*e*^{16}. - Let the common tangent at
*X*meet*AQ*and*PB*at*Y*and*Z*respectively. Then we have ∠*APX*= ∠*AXY*(because*YZ*is tangent to α at*X*) = ∠*ZXB*(because opposite angles are equal) = ∠*PQB*(because*YZ*is tangent to β at*X*). It follows that*AP*is parallel to*QB*, as required. - Let
*a*_{3}*X*^{3}+*a*_{2}*X*^{2}+*a*_{1}*X*+*a*_{0}*I*denote the minimum polynomial of*A*, where*a*_{i}∈**C**and at least one of the*a*_{i}is 1. It is the unique monic polynomial*f*(*X*) of minimal degree such that*f*(*A*) = 0. Since*AB*= 0, we see that*A*is not invertible, hence 0 is an eigenvalue of*A*, consequently*a*_{0}= 0. Set*D*=*a*_{3}*A*^{2}+*a*_{2}*A*+*a*_{1}*I*. Then*D*≠ 0, because*a*_{3}*X*^{3}+*a*_{2}*X*^{2}+*a*_{1}*X*is the minimum polynomial of*A*. Also*AD*=*DA*=*a*_{3}*A*^{3}+*a*_{2}*A*^{2}+*a*_{1}*A*+*a*_{0}*I*= 0 and the result is proven. - Suppose
*n*^{4}-7*n*^{2}+ 1 is a perfect square; we may assume that*n*is positive. We have*n*^{4}-7*n*^{2}+1 = (*n*^{2}-3*n*+ 1)(*n*^{2}+ 3*n*+ 1). Suppose*p*is a prime that divides both*n*^{2}- 3*n*+ 1 and*n*^{2}+ 3*n*+ 1. Then*p*| 6*n*and we see that*p*divides 2,3 or*n*. By inspection, none of these are possible. Thus no prime can divide both*n*^{2}- 3*n*+ 1 and*n*^{2}+ 3*n*+ 1, consequently both these numbers are perfect squares, in particular*n*^{2}+ 3*n*+ 1 is a perfect square. By considering this mod 3, we see that*n*≡0 mod 3, so we may write*n*= 3*m*where*m*is a positive integer. Thus 9*m*^{2}+ 9*m*+ 1 is a perfect square. However (3*m*+ 1)^{2}< 9*m*^{2}+9*m*+ 1 < (3*m*+ 2)^{2}, a contradiction and the result follows. - It is easy to check that
*f*(*x*) =*ax*+*b*satisfies the differential equation for arbitrary constants*a*,*b*; however it cannot satisfy the numerical condition. We need to find another solution to the differential equation. Since the equation is linear, it's worth trying a solution in the form*f*(*x*) =*e*^{rx}, where*r*is a complex constant. Plugging into the differential equation, we obtain*re*^{rx}=*e*^{r(x+1)}-*e*^{rx}, which simplifies to*e*^{r}= 1 +*r*. We need a solution with*r*≠ 0. Write*r*=*a*+*ib*, where*a*,*b*∈**R**. We want to solve*e*^{a+ib}= 1 +*a*+*ib*. By equating the real and imaginary parts, we obtain*e*^{a}cos*b*= 1 + *a**e*^{a}sin*b*= *b*.

Eliminating*a*, we find that*g*(*b*) : = -*b*cos*b*+ (1 + log*b*- log(sin*b*))sin*b*= 0. Since lim_{x-> 0+}*x*log*x*= 0, we see that lim_{b-> 2π+}(log(sin*b*))sin*b*= 0 = lim_{b-> 3π-}(log(sin*b*))sin*b*. ThereforelimSince_{b-> 2π+}*g*(*b*) = -2π < 0 and lim_{b-> 3π-}*g*(*b*) = 3π > 0.*g*is continuous on (2π, 3π), we see that there exists*q*∈(2π, 3π) such that*g*(*q*) = 0. Set*p*= log*q*- log(sin*q*). Then*e*^{p+iq}+*p*+*iq*= 1. Since the differential equation is linear, the real and imaginary parts of*f*will also satisfy the differential equation, and in particular*f*(*x*) : =*e*^{px}sin*qx*will satisfy the differential equation. A routine calculation shows that*f''*(0) = 2*pq*≠ 0, and so the answer to the question is ``yes".**Remark**It can be shown that*p*≈2.09 and*q*≈7.46.

Peter Linnell 2009-11-01