Solved and Unsolved Problems Solved and Unsolved Problems

(Andrei Jaikin-Zapirain, Departamento de Matemáticas, Universidad Autónoma de Madrid & Instituto de Ciencias Matemáticas, CSIC-UAM-UC3M-UCM, Spain, and Dmitri Piontkovski, Faculty of Economic Sciences, Moscow Higher School of Economics, Russia) 230. We are trying to hang a picture on a wall. The picture has a piece of string attached to it forming a loop, and there are 3 nails in the wall that we can wrap the string around. We want to hang the picture so that it does not fall down, but it will upon removal of any of the 3 nails.

Show the following relation: where each product a j b * j ( j = 1, . . . , n) is a rank-one matrix in M n . (T. Ando, Hokkaido University, Sapporo, Japan) 227. Let p and q be two distinct primes with q > p and G a group of exponent q for which the map f p : G → G defined by f p (x) = x p , for all x ∈ G, is an endomorphism. Show that G is an abelian group.
(Dorin Andrica and George CȃtȃlinŢurcaş, Babeş-Bolyai University, Cluj-Napoca, Romania) 228. Let (G, ·) be a group with the property that there is an integer n ≥ 1 such that the map f n : G → G, f n (x) = x n is injective and the map f n+1 : G → G, f n+1 (x) = x n+1 is a surjective endomorphism. Prove that G is an abelian group.
(Dorin Andrica and George CȃtȃlinŢurcaş, Babeş-Bolyai University, Cluj-Napoca, Romania) 229. Let A and B ∈ Mat k (K) be two matrices over a field K. We say that A and B are similar if there exists an invertible matrix C ∈ GL k (K) such that B = C −1 AC. Let A and B ∈ GL k (Q) be two similar invertible matrices over the field of rational numbers Q. Assume that for some integer l, A l+1 B = BA l . Then A and B are the identity matrices. 230. We are trying to hang a picture on a wall. The picture has a piece of string attached to it forming a loop, and there are 3 nails in the wall that we can wrap the string around. We want to hang the picture so that it does not fall down, but it will upon removal of any of the 3 nails.
(Dawid Kielak, Mathematical Institute, University of Oxford, UK) 231. Given a natural number n and a field k, let M n (k) be the full n × n matrix algebra over k. A matrix (a i j ) ∈ M n (k) is said to be centrosymmetric if Let C n (k) denote the set of all centrosymmetric matrices in M n (k). Then C n (k) is a subalgebra of M n (k), called centrosymmetric matrix algebra over k of degree n. Centrosymmentric matrices have a long history (see [1,5]) and applications in many areas, such as in Markov processes, engineering problems and quantum physics (see [2,3,4,6]). In the representation theory of algebras, a fundamental problem for a finite-dimensional algebra is to know if it has finitely many nonisomorphic indecomposable modules (or in other terminology, representations). In our case, the concrete problem on C n (k) reads as follows. Does C n (k) have finitely many nonisomorphic indecomposable modules? If yes, what is the number?
An additional interesting problem (not intimately connected to algebra). Intervals of monotonic changes in the polynomial are located between the roots of its derivative. A derivative of a polynomial is also a polynomial, although of a lesser degree. Using these considerations, construct an algorithm for calculating the real roots of the quadratic equation. Improve it to calculate the real roots of the polynomial of the third, fourth and generally arbitrary degree.
(Igor Kostin, Moscow, Russian Federation) Consider the group ring A of a free finitely generated group (i.e., noncommutative Laurent polynomials) with coefficients in a field k of characteristic zero. Denote by τ : A → k the "trace map" given by the coefficient of monomial 1 ().
Theorem For any a ∈ A the following series is algebraic: The rationale for the minus sign is that if we replace algebra A by Mat(N × N, k) and τ by the usual trace, the resulting series In 2007 I found a ridiculous proof in three steps (basically no progress afterwards): (1) It is known that F := G /G = − n≥1 τ(a n )t n−1 is algebraic (see, e.g., Corollary 6.7.2 in [10], I learned about it from a paper [6] where it was rediscovered). This is a corollary of the theory of algebraic formal languages developed by N. Chomsky and M. Schützenberger in the 60s [4]. (2) Assume that all coefficients of a are integers (it is not a severe restriction), then it is easy to see that coefficients of G are integers (it follows solely from the fact that the free group is torsionfree). , [2] based on ideas of D. V. and G. V. Chudnovsky [5], or by later results of J.-B. Bost [3], series G is also algebraic.
If we replace the free group by a finite group Γ, the resulting series G is again algebraic, a fractional power of a polynomial. This follows from the fact that after an extension of scalars k ⊃ k, the group ring of the finite group is a direct sum of matrix algebras, and that the canonical trace on the group algebra is proportional to the matrix trace on each matrix algebra summand Mat( For a finitely-generated free abelian group, the series F can be calculated by residue formula, and is holonomic, i.e., it satisfies a non-trivial algebraic linear differential equation. Neveretheless, the series G in this case is typically not holonomic, in particular not algebraic. For nonhyperbolic groups the situation is much more tricky, see [7] where it was shown that the analog of generating series F can already be nonholonomic for the arithmetic group S L(4, Z). 232*. Find a purely combinatorial proof of algebraicity of G, not based on results from number theory. In particular, what is an adequate upper bound on the degree of the equation satisfied by G? Is it true that the function G considered as a multi-valued function is either 1 − ct for some constant c ∈ k, or it does not attain zero value for t 0? Also, is it true that the values of all branches of G at t = 0 belong to the set {0, 1} ⊂ P 1 ? 1. Let I be a finite set and let (m i j ) (i, j)∈I×I be a symmetric matrix whose diagonal entries are 1 and whose nondiagonal entries are integers ≥ 2 or ∞. Let W be the group with generators {s i ; i ∈ I} and relations (s i s j ) m i j = 1 for any i, j such that m i j < ∞; this is a Coxeter group. (Examples of Coxeter groups are the Weyl groups of simple Lie algebras; these are finite groups. Other examples are the affine Weyl groups which are almost finite.) For w ∈ W let |w| be the smallest integer n ≥ 0 such that w is a product of n generators s i , i ∈ I. We assume that we are given a weight function L : W → N that is a function such that L(w) > 0 for all w ∈ W − {1} and

II (B) Open problems on Iwahori-Hecke algebras
for i ∈ I and T w T w = T ww for any w, w in W such that |ww | = |w| + |w |; this is the Iwahori-Hecke algebra associated to W, L.
For c ∈ C − {0} let H c = C ⊗ AH where C is viewed as an Aalgebra via the ring homomorphism A → C, v → c. Now H c is also referred to as an Iwahori-Hecke algebra. 233*. Show that the algebras associated in [10] to a supercuspidal representation of a parabolic subgroup of a p-adic reductive group are (up to extension by a group algebra of a small finite group) of the form H q where q is a power of p, with H associated to an affine Weyl group W and with L in the collection Σ W of weight functions on W described in [4, §17], [5], [6].
The statement analogous to (233*) for groups with connected centre over a finite field F q instead of p-adic groups is known to hold, without the words in parenthesis; in that case, W is a Weyl group and Σ W consists of the weight functions on W described in [3, p. 35].
2. There is a unique group homomorphism¯: 234*. Show that there exists an integer N ≥ 0 such that for any (See [7, 13.4]) If W is finite this is obvious. If W is an affine Weyl group, this is known.
We will now assume that (234*) holds. With N as in (234*), we for any x, y, z in W. It follows that for any z ∈ W there is a unique integer a(z) ≥ 0 such that for all x, y in W and for some x, y in W. Hence for x, y, z in W there is a well-defined integer γ x,y,z −1 such that Let J be the free abelian group with basis {t w ; w ∈ W}. For x, y in W we set (This is a finite sum.) 235*. Show that this defines an (associative) ring structure on J (without 1 in general).
Assume now that W is a Weyl group or an affine Weyl group and L = ||. In this case, (235*) is known to be true and the ring J does have a unit element.
More generally, assume that W is an affine Weyl group and L ∈ Σ W (see (233*)); in this case there is a (conjectural) geometric description [8, 3.11], of the elements c w . From this one should be able to deduce (235*) as well as the well-definedness of the Calgebra homomorphism H q → J in [7, 18.9], where H q is as in (233*) and J = C ⊗ J is independent of q. One should expect that the irreducible (finite dimensional) J-modules, when viewed as H qmodules, form a basis of the Grothendieck group of H q -modules. (This is indeed so if L = ||.) This should provide a construction of the "standard modules" of H q which, unlike the construction in [5], [6], does not involve the geometry of the dual group.
3. Assume that W is finite and that L = ||. Let C be a conjugacy class in W; let C min be the set of all w ∈ C such that |w| is minimal. For w ∈ C let N w ∈ A be the trace of the A-linear map H → H, We have N w ∈ Z[v 2 ]. (Note that N w | v=1 is the order of the centralizer of w in W.) From [1] one can deduce that for w ∈ C min , N w depends only on C, not on w. We say that C is positive if N w ∈ N[v 2 ]. For example, if C is an elliptic regular conjugacy class (in the sense of [11]) then C is positive (see [9]). If W is of type A n , the positive conjugacy classes are {1} and the class of the Coxeter element. In the case where W is a Weyl group of exceptional type, a complete list of positive conjugacy classes in W is given in [9]. 236*. Make a list of all positive conjugacy classes in W, assuming that W is a Weyl group of type B n or D n .

218.
Determine the sum of the series where ϕ is the Euler's totient function.
(Dorin Andrica, Babeş-Bolyai University, Cluj-Napoca, Romania) Solution by the proposer. Let (a n )n ≥ 1 be a sequence of real numbers. From the equality x n 1 − x n = x n + x 2n + · · · + x kn + · · · , |x| < 1 we derive ∞ n=1 a n x n where A n = d|n a d and assuming that the power series in the right-hand side of (1) is convergent for |x| < 1. This is the main idea of the so-called Lambert series with the related identities. Now, using the well-known Gauss' identity d|n ϕ(d) = n, the formula (1) yields O(x log log x). (v) Using (iv), prove that ω(n) has normal order log log n, i.e., for every ε > 0, In what follows, the letters p, p 1 , p 2 denote primes.
(i) Using the definition of ω(n) and Mertens' theorem, we obtain (ii) As above, using the definition of ω(n) and Mertens' theorem, we obtain and, again by Mertens' theorem, (iii) Expanding the square and using parts (i), (ii), we deduce that (iv) First we relate the summand (ω(n) − log log n) 2 to the summand (ω(n) − log log x) 2 and then we use (iii): We split the remaining sum over n ≤ x into sums over n ≤ √ x and √ x < n ≤ x, which we bound trivially.
(v) Let ε > 0 and, for any y < x, denote by N(y, x) the number of natural numbers y < n ≤ x for which |ω(n) − log log n| ≥ ε log log n.
Using elementary observations and part (iv), we obtain We leave the proof of (vi) as a challenge to the reader.

Also solved by Mihaly Bencze (Romania) and Socratis Varelogiannis (France).
220. Using Chebyshev's Theorem, prove that for any integer M there exists an even integer 2k such that there are at least M primes p with p + 2k also prime. Unfortunately 2k will depend on M. If it did not, we would have solved the Twin Prime Conjecture, namely, there are infinitely many primes p such that p + 2 is also prime.

(Steven J. Miller, Department of Mathematics & Statistics, Williams College, Massachusetts, USA)
Solution by the proposer. By Chebyshev's theorem, there exist explicit positive constants A and B such that, for x > 30: Ignoring the lone even prime 2, the number of positive differences between the odd primes at most x is π(x)−1 2 , or Looking at the lower and upper bounds for π(x), we get that the number of these differences is essentially at least A 2 x 2 / log 2 x and basically at most B 2 x 2 / log 2 x; however, there are only about x/2 odd numbers which can be these differences.
Thus by the pigeonhole principle, at least one of these positive odd differences must occur at least the average number of times, and thus there is a difference that occurs essentially at least Proof. We have We now begin the solution proper. Let a, b, c as in the problem. We may assume without loss of generality that gcd(a, b) = 1 and that c is squarefree. Indeed, if these conditions are not met, we can replace a by a gcd(a,b) , replace b by b gcd(a,b) and replace c by its largest squarefree divisor, so that this modified setup does satisfy the conditions. Any integer m that satisfies the required conditions in this modified setup will automatically satisfy it in the original setup.
Next, define Q = c/ gcd(b, c); note that Q is squarefree and hence p|Q 2 = d|Q 1. Let X = 2 2 2002 Q Now, using 1 − 1 p ≥ 1 2 , putting in the bounds from the lemma (with k = 2000), and substituting the value of X, we obtain that 1≤m≤X gcd(a+mb,Q)=1 It follows that there exists some m (between 1 and X), such that gcd a + mb, c/ gcd(b, c) = 1.
However, since gcd(a, b) = 1, this implies that gcd(a + mb, c) = 1. This completes the solution of the problem.

Show that
Setting δ = 1/2, deduce the values of ζ(2) and ζ(4). We use Fourier analysis on T; in particular, for n ∈ Z \ {0}, by definition and computation we have Parseval's identity n∈Z | f (n)| 2 = T | f (x)| 2 dx, which is easily verified, then yields which is a rearrangement of our first identity. For the third identity, we use the convolution identity f * f (n) = f (n) 2 and Parseval to write where, by definition of convolution followed by a simple computation (where we use that δ ≤ 1/2), and so ( ) gives δ 4 + 2 ∞ n=1 sin 4 (πδn) π 4 n 4 = 2 3 δ 3 .
Rearranging, we have the third identity.
Comment. The integrals computed above all have an additive combinatorial interpretation: in the case of the fourth-power denominators, the integral measures how many solutions there are to a 1 + a 2 = a 3 + a 4 in the interval [− δ 2 , δ 2 ]. In the case of third-powers, the equation is a 1 + a 2 + a 3 = 0, and in the case of squares it is simply The reason for restricting to δ ≤ 1/2 in two of the cases is that these solutions are easy to measure provided there is no 'wrap-around' when adding two variables.