34th VTRMC, 2012, Solutions

1. Let I denote the value of the integral. We make the substitution y = π/2 -x. Then dx = -dy, and as x goes from 0 to π/2, y goes from π/2 to 0. Also sin(π/2 -x) = cos x and cos(π/2 -x) = sin x. Thus

I = ∫0π/2(sin4x + sin x cos3x + sin2x cos2x + sin3x cos x)/(sin4x + cos4x + 2 sin x cos3x + 2 sin2x cos2x + 2 sin3x cos xdx.

Adding the above to the given integral, we obtain 2I = ∫0π/2dx. Therefore I = π/4.

2. We necessarily have x≥ - 2. Also the left hand side becomes negative for x≥2. Therefore we may assume that x = 2 cos t for 0≤t≤π. After making this substitution, the equation becomes cos 3t + cos(t/2) = 0. Using a standard trig formula ( 2 cos A cos B = cos(A + B) + cos(A -B)), this becomes cos(7t/4)cos(5t/4) = 0. This results in the solutions t = 2π/5, 2π/7, 6π/7. Therefore x = 2 cos(2π/5), 2 cos(2π/7), 2 cos(6π/7).

3. We make a, b, c, d, e be the roots of the quintic equation x5 + px4 + qx3 + rx2 + sx + t = 0. Using the first and last equations, we get p = t = 1. Let Z = {a, b, c, d, e}. Then

2q = ∑u, v∈Z, u≠vuv = (a + ... + e)2 - (a2 + ... + e2) = -14,

so q = -7. Next s = abcde(1/a + ... + 1/e) = -1/ - 1 = 1. Finally

 r = abcde(∑u, v∈Z, u≠vuv) = abcde((1/a + ... +1/e)2 - (1/a2 + ... +1/e2)) = -14,

so r = -7.

Similarly s = 1 and r = -7. Therefore a, b, c, d, e are the roots of x5 + x4 -7x3 -7x2 + x + 1 = 0. By inspection, -1 is a root and the equation factors as

(x + 1)(x4 -7x2 +1) = (x + 1)(x2 -3x + 1)(x2 + 3x + 1).

Using the quadratic formula, it follows that a, b, c, d, e are (in whatever order you like)

-1,(±3±√5)/2.

4. We repeatedly use the fact that if n is a positive integer and a∈ℤ is prime to n, then aφ(n)≡1 mod n where φ is Euler's totient function.

We first show that f (n)≡3 mod 4 for all n≥1. We certainly have f (1)≡3 mod 4. Since f (n) is always odd, we see that f (n + 1)≡3f(n)≡3f(n-1)f (n) mod 4 and we deduce that f (n)≡3 mod 4 for all n≥1.

Now we show that f (n)≡f (3) mod 25 for all n≥3. First observe that f (n + 1)≡3f(n)≡3f(n-1)f (n) mod 5 for n≥2, provided f (n) = f (n - 1) mod 4, which is true by the previous paragraph. It follows that f (n + 1)≡f (n) mod 20 for all n≥2. Therefore f (n + 1)≡3f(n)≡3f(n-1)f (n) mod 25, provided n≥3, and our assertion is proven. Since the last two digits of f (3) are 87, the last two digits of f (2012) are also 87.

5. Let f (n) = 1/(ln n) - (1/ln n)(n+1)/n. Then f (n)ln n = 1 - (ln n)-1/n. Assume that n > 27. Since ln n > e, we see that f (n)ln n > 1 -e-1/n. Therefore nf (n)ln n > n(1 -e-1/n). By L'hôpital's rule, limn→∞n(1 -e-1/n) = 1. Therefore nf (n)ln n > 1/2 for sufficiently large n, so f (n) > 1/(2n ln n). Since n=21/(n ln n) is divergent, it follows that n=2f (n) is also divergent.

6. We shall prove by induction that an = p if p is a prime and n = pm for some positive integer m, and 1 otherwise. This is clear in the case n = pm, because then there are exactly m - 1 nontrivial divisors of pm, and each contributes p to the denominator of the displayed fraction. The case n = pq, where p, q are distinct primes, is also clear, because then p and q are the only nontrivial divisors of n, and they contribute p and q respectively to the denominator.

Now assume that n is neither a prime power, nor a product of two distinct primes, and assume the result is true for all smaller values of n. Then we may write n = pm, where p is a prime and m is not a prime power. Write m = pak, where a is a nonnegative integer and k is prime to p. If d | n, then either d | m or d = pa+1r, where r | k. Note that in the latter case, d is a prime power only when r = 1. Therefore

an = amp/apa+1 = 1p/p = 1.

by induction, which proves the claim. It follows that a999000 = 1.

7. Let 0 and I denote the zero and identity 2×2 matrices respectively. Let A denote one of the three matrices. The result is clear if A = 0 or I, because every matrix commutes with 0 and I. Next note that Ai = Aj, where 0 < i < j < 5, and we see that the minimum polynomial of A divides xj -xi.

Suppose 0 is not an eigenvalue of A. Then A is invertible and it follows that Aj-i = I, in particular I is one of the matrices. Since I commutes with all matrices, the result follows in this case.

Thus we may assume that 0 is an eigenvalue of A. Next suppose both the eigenvalues of A are 0 (i.e. A has a repeated eigenvalue 0). Then the Jordan canonical form of A is either 0 or

(
 0 1 0 0
)

In both case, A2 = 0, and since 0 commutes with all matrices, the result follows in this case.

Therefore we may assume that A has one eigenvalue 0 and another eigenvalue λ≠ 0. Since the minimum polynomial of A divides xj -xi where 0 < i < j < 5, we see that the possibilities for λ are 1, -1, or ω where ω is a primitive cube root of 1. Since the eigenvalues of A are distinct, it is diagonalizable and in particular, its Jordan canonical form will be

(
 λ 0 0 0
)

If λ = ω, then {A, A2, A3} are three distinct commuting matrices, and the result is proven in this case. Thus we may assume that the Jordan canonical form for A is

(
 ± 1 0 0 0
)

Now not all the Ai can have Jordan canonical form

(
 1 0 0 0
)

because then tr(A1 + A2 + A3) = 3, so at least one of the matrices, say A1, has trace -1. It should be pointed out that we may assume that the Ai are distinct, if not, then the three matrices come from {A1, A12}, and since A1 commutes with A12, the result follows in this case.

Suppose tr(A2) = -1 and tr(A3) = 1. Then A12 = A22 = A3 and A3 commutes with A1 and A2, and the result is proven in this case.

Finally suppose tr(A2) = tr(A3) = 1. Then without loss of generality, we may assume that A12 = A2, and so A1 = -A2. Thus -A3A1 or A2. Since A2A3 = -A1A3, we see that A2A3 = A1 or A2. Similarly A3A2 = A1 or A2. Since tr(A2A3) = tr(A3A2), we deduce that A2A3 = A3A2. Also A1A2 = A2A1, and the result follows.

Peter Linnell 2012-11-03