38th VTRMC, 2016, Solutions
I = ∫12(ln x)/(2 - 2x + x2) dx.
We make the substitution y = 2/x. Then
dx = -2y-2dy and we
I = ∫21(-2y-2ln(2/y))/(2 - 4/y - 4/y2) dy = ∫12(ln 2 - ln y)/(y2 -2y + 2) dy.
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.
an = (2n)!/(4nn!n!). Then
an/an-1 = (2n - 1)/(2n) = 1 - 1/(2n). Therefore
(n - 1)/n≤(an/an-1)2≤n/(n + 1)
Now if bn = 1/n, then
1/4n≤an2≤1/(n + 1) and hence
1/((4n)k/2)≤an≤1/(n + 1)k/2.
∑1/nk/2 is convergent if and only if k > 2, we deduce
that the series is convergent for k > 2 and divergent for
- Let I denote the identity matrix in
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
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
Y≠Y' otherwise Y
would have two rows equal and therefore would not be invertible).
Z2≠I, then we can pair
it with Z-1 and we see that the number of matrices satisfying
GLn(ℤ2) is even. Therefore the number
of matrices satisfying X2 = I is even and the result follows.
- 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.
- 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.
| ||MN =
I + A + B + AB -XY
| ||PY -YB
I + P + Q + PQ -YX
Therefore NM = I and in particular
I + A + B + BA -XY = I. The
- Proceed by induction on k. Let ck denote the constant term of
fk. For the base case k = 1, we need only
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
||= (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).
ak(±1) to ak(0), we consider
||= ∏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).
(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).
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.