38th VTRMC, 2016, Solutions

1. Write I = ∫12(ln x)/(2 - 2x + x2dx. We make the substitution y = 2/x. Then dx = -2y-2dy and we have

I = ∫21(-2y-2ln(2/y))/(2 - 4/y - 4/y2dy = ∫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 an = (2n)!/(4nn!n!). Then an/an-1 = (2n - 1)/(2n) = 1 - 1/(2n). Therefore

(n - 1)/n≤(an/an-1)2n/(n + 1)

for all n∈ℕ. Now if bn = 1/n, then

bn/bn-1≤(an/an-1)2bn+1/bn.

Therefore 1/4nan2≤1/(n + 1) and hence

1/((4n)k/2)≤an≤1/(n + 1)k/2.

Since ∑1/nk/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 Mn(ℤ2). If A∈Mn(ℤ2) and A2 = 0, then (I + A)2 = I + 2A + A2 = I because we are working mod 2, and we see that I + A∈GLn(ℤ2), the invertible matrices in Mn(ℤ2). Conversely if X∈GLn(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 GLn(ℤ2) is even (if Y∈GLn(ℤ2), then we can pair it with the matrix Y' obtained from Y by interchanging the first two rows of Y, and note that YY' otherwise Y would have two rows equal and therefore would not be invertible). Now if Z∈GLn(ℤ2) and Z2I, then we can pair it with Z-1 and we see that the number of matrices satisfying Z2I in GLn(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 ap -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 aS 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 = 12r-1 = -1. Therefore m(m + 2) = 3n2. If p≠2, 3 is a prime and pa is the largest power of p dividing n, then p2a is the largest power of p dividing 3n2. Since p cannot divide both m and m + 2, we see that either p ∤ m or p2a | 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 = 22r-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 ck denote the constant term of fk. 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

ck+1 = ((1 -q2k+1)(1 -q2k+2))/((1 -qk+1)2)ck = ((1 -q2k+1)(1 + qk+1))/(1 -qk+1)ck.

We will prove that the constant term of fk(X) satisfies the same recurrence relation, which gives the induction step. Let ak(i) denote the coefficient of Xi in fk. From

 fk+1(X) = (1 -qkX)(1 -qk+1X-1)fk(X) = (1 -qkX -qk+1X-1 + q2k+1)fk(X)

we deduce that

ak+1(0) = (1 + q2k+1)ak(0) -qkak(-1) -qk+1ak(1).

We want a recurrence relation for ak(0). To relate ak(±1) to ak(0), we consider

 fk(qX) = ∏i=0k-1((1 -qi+1X)(1 -qiX-1)) = ((1 -qkX)(1 -X-1))/((1 -X)(1 -qkX-1))fk(X) = (1 -qkX)/(qk -X)fk(X).

Hence (qk -X)fk(qX) = (1 -qkX)fk(X). Equating coefficients of X0 and X1 on both sides, we obtain

ak(-1) = q(qk -1)ak(0)/(1 -qk+1),        ak(1) = (qk -1)ak(0)/(1 -qk+1).

Therefore

ak+1(0) = (1 + q2k+1 -2qk+1(qk -1)/(1 -qk+1))ak(0) = (1 -q2k+1)(1 + qk+1)ak(0)/(1 -qk+1)

and this completes the proof.

Peter Linnell 2016-10-30