38th VTRMC, 2016, Solutions

1. Write I = ∫₁²(ln x)/(2 - 2x + x²) dx. We make the substitution y = 2/x. Then dx = -2y^-2dy and we have
   I = ∫₂¹(-2y^-2ln(2/y))/(2 - 4/y - 4/y²) dy = ∫₁²(ln 2 - ln y)/(y² -2y + 2) dy.
   Therefore
   2I = ∫₁²(ln 2)/(y² -2y + 2) dy = ∫₀¹(ln 2)/(x² +1) dx
   by making the substitution x = y - 1. We conclude that I = (πln 2)/8.

2. Set a_n = (2n)!/(4ⁿn!n!). Then a_n/a_n-1 = (2n - 1)/(2n) = 1 - 1/(2n). Therefore
   (n - 1)/n≤(a_n/a_n-1)²≤n/(n + 1)
   for all 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.
   Therefore 1/4n≤a_n²≤1/(n + 1) and hence
   1/((4n)^k/2)≤a_n≤1/(n + 1)^k/2.
   Since ∑1/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.

3. Let I denote the identity matrix in M_n(ℤ₂). If A∈M_n(ℤ₂) and A² = 0, then (I + A)² = I + 2A + A² = I because we are working mod 2, and we see that I + A∈GL_n(ℤ₂), the invertible matrices in M_n(ℤ₂). Conversely if X∈GL_n(ℤ₂), and X² = I, then (I + X)² = 0. We deduce that the number of matrices A satisfying A² = 0 is precisely the number of matrices satisfying X² = I. Since n≥2, the number of matrices in GL_n(ℤ₂) is even (if Y∈GL_n(ℤ₂), 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(ℤ₂) and Z²≠I, then we can pair it with Z^-1 and we see that the number of matrices satisfying Z²≠I in GL_n(ℤ₂) is even. Therefore the number of matrices satisfying X² = I is even and the result follows.

4. First observe that if p > 2 is a prime and a < p is such that a² + 1 is divisible by p, then a≠p -a and P(a) = P(p -a) = p. Indeed a² + 1 and (p -a)² +1 = (a² + 1) + p(p - 2a) are divisible by p and are smaller than p², 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 a² + 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 2p, 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 2pq and thus (b + p)² +1≥2pq. This means q < b + p, otherwise (b + p)² +1≤(2p - 1)q + 1 (because b < p) < 2pq. 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.

5. The equality yields 1 + m -n√3 = (2 - √3)^2r-1 and hence (1 + m)² -3n² = 1^2r-1 = -1. Therefore m(m + 2) = 3n². 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 3n². 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 3n², 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.

6. 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.

7. 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 f₁(X) = (1 -X)(1 -qX^-1) = 1 + q -X -qX^-1 and c₁ = (1 -q²)/(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.
   We will prove that the constant term of f_k(X) satisfies the same recurrence relation, which gives the induction step. Let a_k⁽ⁱ⁾ denote the coefficient of Xⁱ in f_k. From

   f_k+1(X) = (1 -q^kX)(1 -q^k+1X^-1)f_k(X) = (1 -q^kX -q^k+1X^-1 + q^2k+1)f_k(X)

   
   
   we deduce that
   a_k+1⁽⁰⁾ = (1 + q^2k+1)a_k⁽⁰⁾ -q^ka_k^(-1) -q^k+1a_k⁽¹⁾.
   We want a recurrence relation for a_k⁽⁰⁾. To relate a_k^(±1) to a_k⁽⁰⁾, we consider

   f_k(qX) = ∏_i=0^k-1((1 -qⁱ⁺¹X)(1 -qⁱX^-1)) = ((1 -q^kX)(1 -X^-1))/((1 -X)(1 -q^kX^-1))f_k(X) = (1 -q^kX)/(q^k -X)f_k(X).

   
   
   Hence (q^k -X)f_k(qX) = (1 -q^kX)f_k(X). Equating coefficients of X⁰ and X¹ 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).
   Therefore
   a_k+1⁽⁰⁾ = (1 + q^2k+1 -2q^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)
   and this completes the proof.