- Write
*I*= ∫_{1}^{4}(*x*- 2)/((*x*^{2}+4)√*x*)*dx*and make the substitution*y*= √*x*. Then*dx*= 2*y**dy*and*I*= ∫_{1}^{2}(2*y*^{2}-4)/(*y*^{4}+4)*dy*. Now*y*^{4}+4 = (*y*^{2}-2*y*+ 2)(*y*^{2}+ 2*y*+ 2) and using partial fractions, we find(2It follows that 2*y*^{2}-4)/(*y*^{4}+4) = (*y*- 1)/(*y*^{2}-2*y*+ 2) - (*y*+ 1)/(*y*^{2}+ 2*y*+ 2).*I*= [ln(*y*^{2}-2*y*+ 2) - ln(*y*^{2}+2*y*+ 2)]_{1}^{2}= ln 2 - ln 1 - (ln 10 - ln 5) = 0, so the answer is 0. - The first few terms (starting with
*a*_{-1}) are -1, 0, 3, 8, so it seems reasonable that*a*_{n}=*n*^{2}- 1; let us prove this by induction on*n*. The result is certainly true if*n*= 0 or 1. So suppose*a*_{r}=*r*^{2}- 1 for*r*≤*n*. Then*a*_{n+1}= ( *n*^{2}-1)^{2}- (*n*+ 1)^{2}((*n*- 1)^{2}- 1) - 1= *n*^{4}-2*n*^{2}+1 - (*n*^{4}-2*n*^{2}+1) +*n*^{2}+ 2*n*+ 1 - 1= *n*^{2}+2*n*= (*n*+ 1)^{2}- 1.

and the induction is complete. Thus*a*_{n}=*n*^{2}- 1 for all*n*∈**N**and we deduce that*a*_{100}= 9999. - Let
*S*= ∑_{k=1}^{n}(*k*^{2}- 2)/((*k*+ 2)!) where*n*∈**N**. We have(By telescoping series, we find that*k*^{2}- 2)/((*k*+ 2)!) = 1/*k*! - 1/(*k*+ 2)! - 3(1/(*k*+ 1)! - 1/(*k*+ 2)!).*S*= 1 + 1/2 - 3/2 - 1/(*n*+ 1)! - 1/(*n*+ 2)! + 3/(*n*+ 2)! = -1/(*n*+ 1)! + 2/(*n*+ 2)!. Since lim_{n-> ∞}1/*n*! = 0, it follows that the required sum is 0. - We repeatedly use the Chinese remainder theorem without further
comment. Let
*b*∈**Z**. We claim that |{[*br*] |*r*∈**Z**}| =*mn*/(*mn*,*b*). Set*k*= (*mn*,*b*). Then {[*kr*] |*r*∈**Z**} = {[*kr*] |*r*= 1, 2,...,*mn*/*k*} so |{[*kr*] |*r*∈**Z**}| =*mn*/*k*. It follows that |{[*br*] |*r*∈**Z**}|≤*mn*/*k*. On the other hand there exists*c*∈**Z**such that and*bc*≡*k*mod*mn*. We conclude that |{[*br*] |*r*∈**Z**}|≥*m*and the claim is established.Therefore

*mn*/(*a*,*mn*) =*m*. Thus (*a*,*mn*) =*n*, hence*n*|*a*and we may write*a*=*sn*where*s*∈**Z**; clearly (*s*,*m*) = 1. Now if*t*∈**Z**, then (*s*+*tm*)*n*≡*a*mod*mn*, i.e. [*n*(*s*+*tm*)] = [*a*]. Since (*s*,*m*) = 1, we may choose*t*so that (*s*+*tm*,*mn*) = 1. The result follows by setting*q*=*s*+*tm*. - Set
*f*(*x*) =*x*^{1+1/x}-*x*- ln*x*. We first show that lim_{x-> ∞}*f*(*x*) = 0. Note that lim_{x-> ∞}*x*^{1/x}= 1 because lim_{x-> ∞}ln(*x*^{1/x}) = lim_{x-> ∞}(ln*x*)/*x*= 0. Thus lim_{x-> ∞}*f*(*x*)/*x*= 0. Now we apply l'hôpital's rule. We obtainlim _{x-> ∞}*f*(*x*)= lim _{x-> ∞}(*f*(*x*)/*x*)/(1/*x*)= lim _{x-> ∞}((*f*(*x*)/*x*)')/(- 1/*x*^{2})= lim _{x-> ∞}-*x*^{2}(*x*^{1/x}-1 - (ln*x*)/*x*)'= lim _{x-> ∞}-*x*^{2}(*x*^{1/x}/*x*^{2}- (*x*^{1/x}ln*x*)/*x*^{2}-1/*x*^{2}+ (ln*x*)/*x*^{2})= lim _{x-> ∞}(*x*^{1/x}-1)(ln*x*- 1).

Thus we need to prove lim_{x-> ∞}(*x*^{1/x}-1)ln*x*= 0. Again we apply l'hôpital's rule.lim _{x-> ∞}(*x*^{1/x}-1)ln*x*= lim _{x-> ∞}(*x*^{1/x}-1)/(1/ln*x*)= lim _{x-> ∞}(*x*^{1/x}(1 - ln*x*)/*x*^{2})/(- 1/(*x*(ln*x*)^{2}))= lim _{x-> ∞}((ln*x*)^{3})/*x*= 0.

Thus lim_{x-> ∞}*f*(*x*) = 0 and we deduce that lim_{x-> ∞}*f*(2*x*) -*f*(*x*) = 0. But*f*(2*x*) -*f*(*x*)= (2 *x*)^{1+1/(2x)}-2*x*- ln 2*x*-*x*^{1+1/x}+*x*+ ln*x*= (2 *x*)^{1+1/(2x)}-*x*^{1+1/x}-*x*- ln 2

and we deduce that lim_{x-> ∞}(2*x*)^{1+1/(2x)}-*x*^{1+1/x}-*x*= ln 2. - Let < indicate the usual order on
**Q**; thus < is an asymmetric relation on**Q**. Define*T*=*S*×**Q**and let ≺ denote the lexicographic asymmetric relation on*T*, that is (*a*,*p*)≺(*b*,*q*) if and only if*a*<*b*or*a*=*b*and*p*<*q*. It is easily checked that ≺ is asymmetric. Also we may identify*S*with*S*× 0 via*s*|--> (*s*, 0), and then the restriction of ≺ to*S*is <. Now suppose (*a*,*p*)≺(*b*,*q*). If*a*<*b*, then (*a*,*p*)≺(*a*,*p*+ 1)≺(*b*,*q*). On the other hand if*a*=*b*, then*p*<*q*and (*a*,*p*)≺(*a*,(*p*+*q*)/2)≺(*a*,*q*) and the result is proven. - Assume that the roots
*x*_{1},...,*x*_{100}are all real. By the Viete relations we have ∑_{i=1}^{100}*x*_{i}= -20 and ∑_{i < j}*x*_{i}*x*_{j}= 198. Therefore ∑_{i=1}^{100}*x*_{i}^{2}= (- 20)^{2}- 2*198 = 4 and we deduce that (∑_{i=1}^{100}*x*^{i})^{2}= 100∑_{i=1}^{100}*x*_{i}^{2}. Now apply the Cauchy-Schwartz triangle inequality, that is (∑_{i}*x*_{i}*y*_{i})^{2}≤∑_{i}*x*_{i}^{2}∑_{i}*y*_{i}^{2}with equality if and only if one of (*x*_{i}),(*y*_{i}) is a scalar multiple of the other, in particular taking*y*_{i}= 1 for all*i*, we obtain (∑_{i=1}^{100}*x*_{i})^{2}≤100∑_{i=1}^{100}*x*_{i}^{2}with equality if and only if (*x*_{i}) is a scalar multiple of (1,..., 1), i.e. all the*x*_{i}are equal. We deduce that all the*x*_{i}are equal. But the product of the roots is 1, consequently all the*x*_{i}are 1 or all the*x*_{i}are -1, which contradicts the fact that ∑_{i=1}^{100}*x*_{i}= -20.

Peter Linnell 2011-10-31