- Write
*I*= ∫_{1}^{2}(ln*x*)/(2 - 2*x*+*x*^{2})*dx*. We make the substitution*y*= 2/*x*. Then*dx*= -2*y*^{-2}*dy*and we have*I*= ∫_{2}^{1}(-2*y*^{-2}ln(2/*y*))/(2 - 4/*y*- 4/*y*^{2})*dy*= ∫_{1}^{2}(ln 2 - ln*y*)/(*y*^{2}-2*y*+ 2)*dy*.2by making the substitution*I*= ∫_{1}^{2}(ln 2)/(*y*^{2}-2*y*+ 2)*dy*= ∫_{0}^{1}(ln 2)/(*x*^{2}+1)*dx**x*=*y*- 1. We conclude that*I*= (πln 2)/8. - Set
*a*_{n}= (2*n*)!/(4^{n}*n*!*n*!). Then*a*_{n}/*a*_{n-1}= (2*n*- 1)/(2*n*) = 1 - 1/(2*n*). Therefore(for all*n*- 1)/*n*≤(*a*_{n}/*a*_{n-1})^{2}≤*n*/(*n*+ 1)*n*∈ℕ. Now if*b*_{n}= 1/*n*, then*b*_{n}/*b*_{n-1}≤(*a*_{n}/*a*_{n-1})^{2}≤*b*_{n+1}/*b*_{n}.*n*≤*a*_{n}^{2}≤1/(*n*+ 1) and hence1/((4Since ∑1/*n*)^{k/2})≤*a*_{n}≤1/(*n*+ 1)^{k/2}.*n*^{k/2}is convergent if and only if*k*> 2, we deduce that the series is convergent for*k*> 2 and divergent for*k*≤2. - Let
*I*denote the identity matrix in M_{n}(ℤ_{2}). If*A*∈M_{n}(ℤ_{2}) and*A*^{2}= 0, then (*I*+*A*)^{2}=*I*+ 2*A*+*A*^{2}=*I*because we are working mod 2, and we see that*I*+*A*∈GL_{n}(ℤ_{2}), the invertible matrices in M_{n}(ℤ_{2}). Conversely if*X*∈GL_{n}(ℤ_{2}), and*X*^{2}=*I*, then (*I*+*X*)^{2}= 0. We deduce that the number of matrices*A*satisfying*A*^{2}= 0 is precisely the number of matrices satisfying*X*^{2}=*I*. Since*n*≥2, the number of matrices in GL_{n}(ℤ_{2}) is even (if*Y*∈GL_{n}(ℤ_{2}), then we can pair it with the matrix*Y'*obtained from*Y*by interchanging the first two rows of*Y*, and note that*Y*≠*Y'*otherwise*Y*would have two rows equal and therefore would not be invertible). Now if*Z*∈GL_{n}(ℤ_{2}) and*Z*^{2}≠*I*, then we can pair it with*Z*^{-1}and we see that the number of matrices satisfying*Z*^{2}≠*I*in GL_{n}(ℤ_{2}) is even. Therefore the number of matrices satisfying*X*^{2}=*I*is even and the result follows. - First observe that if
*p*> 2 is a prime and*a*<*p*is such that*a*^{2}+ 1 is divisible by*p*, then*a*≠*p*-*a*and*P*(*a*) =*P*(*p*-*a*) =*p*. Indeed*a*^{2}+ 1 and (*p*-*a*)^{2}+1 = (*a*^{2}+ 1) +*p*(*p*- 2*a*) are divisible by*p*and are smaller than*p*^{2}, so they cannot be divisible by any prime greater than*p*.We will prove the stronger statement that there are infinitely many primes

*p*for which*P*(*x*) =*p*has at least three positive integer solutions, so assume by way of contradiction that there are finitely many such primes and let*s*be the maximal prime among these; if there are no solutions, set*s*= 2. Let*S*be the product of all primes not exceeding*s*. If*p*=*P*(*S*), then*p*is coprime to*S*and thus*p*>*s*. Let*a*be the least positive integer such that*a*≡*S*mod*p*. Then*a*^{2}+ 1 is divisible by*p*, hence*P*(*a*) =*P*(*p*-*a*) =*p*because*p*>*a*. Let*b*=*a*if*a*is even, otherwise let*b*=*p*-*a*. Then (*b*+*p*)^{2}+ 1 is divisible by 2*p*, so*P*(*b*+*p*)≥*p*. If*P*(*b*+*p*) =*p*, we arrive at a contradiction. Therefore*P*(*b*+*p*) = :*q*>*p*and (*b*+*p*)^{2}+ 1 is divisible by 2*pq*and thus (*b*+*p*)^{2}+1≥2*pq*. This means*q*<*b*+*p*, otherwise (*b*+*p*)^{2}+1≤(2*p*- 1)*q*+ 1 (because*b*<*p*) < 2*pq*. Now let*c*be the least positive integer such that*c*=*b*+*p*mod*q*. We have*P*(*c*) =*P*(*q*-*c*) =*P*(*b*+*p*) =*q*>*p*>*s*, another contradiction and the proof is finished. - The equality yields
1 +
*m*-*n*√3 = (2 - √3)^{2r-1}and hence (1 +*m*)^{2}-3*n*^{2}= 1^{2r-1}= -1. Therefore*m*(*m*+ 2) = 3*n*^{2}. If*p*≠2, 3 is a prime and*p*^{a}is the largest power of*p*dividing*n*, then*p*^{2a}is the largest power of*p*dividing 3*n*^{2}. Since*p*cannot divide both*m*and*m*+ 2, we see that either*p*∤*m*or*p*^{2a}|*m*, in either case the power of*p*that divides*m*is an even. It remains to prove that the largest power of 2 and 3 that divides*m*is also even. Now if 2 divides*m*, then the largest power of 2 that divides*m*(*m*+ 2), and hence also 3*n*^{2}, is odd which is not possible. All that remains to be proven is that 3 does not divide*m*. However we have 1 +*m*= 2^{2r-1}mod 3, which shows that 3 does not divide*m*as required. - Write
*M*=( *I*+*A*- *X*- *Y**I*+*P*) *N*=( *I*+*B**X**Y**I*+*Q*) Then

*MN*=( *I*+*A*+*B*+*AB*-*XY**AX*-*XQ**PY*-*YB**I*+*P*+*Q*+*PQ*-*YX*) = *I*Therefore

*NM*=*I*and in particular*I*+*A*+*B*+*BA*-*XY*=*I*. The result follows. - Proceed by induction on
*k*. Let*c*_{k}denote the constant term of*f*_{k}. For the base case*k*= 1, we need only observe that*f*_{1}(*X*) = (1 -*X*)(1 -*qX*^{-1}) = 1 +*q*-*X*-*qX*^{-1}and*c*_{1}= (1 -*q*^{2})/(1 -*q*) = 1 +*q*. For any*k*, we have*c*_{k+1}= ((1 -*q*^{2k+1})(1 -*q*^{2k+2}))/((1 -*q*^{k+1})^{2})*c*_{k}= ((1 -*q*^{2k+1})(1 +*q*^{k+1}))/(1 -*q*^{k+1})*c*_{k}.*f*_{k}(*X*) satisfies the same recurrence relation, which gives the induction step. Let*a*_{k}^{(i)}denote the coefficient of*X*^{i}in*f*_{k}. From*f*_{k+1}(*X*)= (1 - *q*^{k}*X*)(1 -*q*^{k+1}*X*^{-1})*f*_{k}(*X*)= (1 - *q*^{k}*X*-*q*^{k+1}*X*^{-1}+*q*^{2k+1})*f*_{k}(*X*)

we deduce that*a*_{k+1}^{(0)}= (1 +*q*^{2k+1})*a*_{k}^{(0)}-*q*^{k}*a*_{k}^{(-1)}-*q*^{k+1}*a*_{k}^{(1)}.*a*_{k}^{(0)}. To relate*a*_{k}^{(±1)}to*a*_{k}^{(0)}, we consider*f*_{k}(*qX*)= ∏ _{i=0}^{k-1}((1 -*q*^{i+1}*X*)(1 -*q*^{i}*X*^{-1}))= ((1 - *q*^{k}*X*)(1 -*X*^{-1}))/((1 -*X*)(1 -*q*^{k}*X*^{-1}))*f*_{k}(*X*)= (1 - *q*^{k}*X*)/(*q*^{k}-*X*)*f*_{k}(*X*).

Hence (*q*^{k}-*X*)*f*_{k}(*qX*) = (1 -*q*^{k}*X*)*f*_{k}(*X*). Equating coefficients of*X*^{0}and*X*^{1}on both sides, we obtain*a*_{k}^{(-1)}=*q*(*q*^{k}-1)*a*_{k}^{(0)}/(1 -*q*^{k+1}),*a*_{k}^{(1)}= (*q*^{k}-1)*a*_{k}^{(0)}/(1 -*q*^{k+1}).*a*_{k+1}^{(0)}= (1 +*q*^{2k+1}-2*q*^{k+1}(*q*^{k}-1)/(1 -*q*^{k+1}))*a*_{k}^{(0)}= (1 -*q*^{2k+1})(1 +*q*^{k+1})*a*_{k}^{(0)}/(1 -*q*^{k+1})

Peter Linnell 2016-10-30