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Problemen NAW 5/14 nr. 3 september 2013

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Redactie:

Johan Bosman Gabriele Dalla Torre Christophe Debry Jinbi Jin Marco Streng Wouter Zomervrucht Problemenrubriek NAW Mathematisch Instituut Universiteit Leiden Postbus 9512 2300 RA Leiden

problems@nieuwarchief.nl www.nieuwarchief.nl/problems

This Problem Section is open to everyone; everybody is encouraged to send in solutions and propose problems. Group contributions are welcome.

For each problem, the most elegant correct solution will be rewarded with a book token worth D20. At times there will be a Star Problem, to which the proposer does not know any solution.

For the first correct solution sent in within one year there is a prize of^D100.

When proposing a problem, please either include a complete solution or indicate that it is intended as a Star Problem. Electronic submissions of problems and solutions are preferred (problems@nieuwarchief.nl).

The deadline for solutions to the problems in this edition is 1 December 2013.

Problem A(proposed by Hendrik Lenstra)

Leta, b ∈ C. Show that if there exists an irreducible polynomialf ∈ Q[X]such thatf (a) = f (a + b) = f (a + 2b) = 0, thenb = 0.

Problem B(The attribution will appear in the March issue of 2014)

Letnbe a positive integer, and lete_ijbe an integer for all1 ≤j ≤ i ≤ n. Show that there exists ann × n-matrix with integer entries such that the eigenvalues of the top lefti × i-minor aree_i1, . . . , e_ii(with multiplicity).

Problem C(proposed by Bart de Smit and Hendrik Lenstra)

LetAbe a finite commutative unital ring. Does there exist a pair(B, f )withBa finite com- mutative unital ring in which every ideal is principal, andf an injective ring homomorphism A → B?

Edition 2013-1 We received solutions from Wouter Cames van Batenburg (Leiden), Hao Chen (Seattle), Charles Delorme (Paris), Florian Eisele (Brussels), Alex Heinis (Hoofddorp), Pieter de Groen (Brussels), Thijmen Krebs (Nootdorp), Guido Senden (Groningen), Sep Thijssen (Ni- jmegen) and Traian Viteam (Punta Arenas).

Problem 2013-1/A Consider a regularn-gonP₁P₂. . . P_n, and drawn − 3diagonals such that there are no intersection points in the interior. The polygon is now divided inton − 2triangles.

Lettibe the number of such triangles that have a vertex atPi. Show that

t1− 1

t2− 1

. . . − 1 t_n−1

= 0.

Solution We received solutions from Charles Delorme, Alex Heinis, Pieter de Groen, Guido Senden, Sep Thijssen, Traian Viteam and Thijmen Krebs. The book token goes to Sep Thijssen.

All solutions followed the same general strategy. The following solution is most similar to that of Charles Delorme, Pieter de Groen and Sep Thijssen.

We use[t₁, . . . , t_n−1]to denote the continued fraction from the problem statement, and we will prove the identity by induction for all convex (not necessarily regular)n-gons withn ≥ 3. Note that the base casen = 3is trivial as1 −¹₁ = 0.

Now assumen ≥ 4and assume that the identity holds forn − 1. The induction step works by finding a triangle with two exterior edges, that is, a trianglePk−1PkPk+1withtk= 1, and by removing that triangle from then-gon, which yields an(n − 1)-gon.

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Lemma 1. Ifn ≥ 4, then there exists an integerkwith2 ≤k ≤ n − 1andt_k= 1.

Proof. We prove first that there are at least two integerskwith1 ≤k ≤ nwithtk= 1. Suppose that there is at most one such integer, that is, suppose that there is at most one triangle with two exterior edges. The total number of exterior edgesnthen is at most2 + (#triangles− 1) = n − 1, which is a contradiction.

Now note that no two such integerskcan correspond to adjacent vertices, so that at least one

of them satisfies2 ≤k ≤ n − 1. _

Now letkbe as in the lemma and assume for nowk 6= n − 1. Consider the(n − 1)-gon with the vertexPkand the edgesPk−1PkandPkPk+1removed. For that(n − 1)-gon, the triangle count tkis omitted, and the adjacent numberstk−1andtk+1are lowered by1. In other words, the induction hypothesis yields

[t1, . . . , tk−2, tk−1− 1, tk+1− 1, tk+2, . . . , tn−1] = 0.

In particular, the proof of the casek 6= n − 1is finished once we prove the following key identity.

Lemma 2. We have

[t1, . . . , tk−2, tk−1 , 1, tk+1 , tk+2, . . . , tn−1]

=[t₁, . . . , tk−2, tk−1− 1, tk+1− 1, tk+2, . . . , tn−1].

Proof. From the definition, we directly get

[a₁, . . . , am, [b₁, . . . , bn]] = [a₁, . . . , am, b₁, . . . , bn]. (1)

Moreover, it is a simple matter of writing things out to see

[a, 1, b] = [a − 1, b − 1]. (2)

Indeed, we have

a − 1

1 −_b¹ =a − b

b − 1=a − 1 − 1 b − 1.

The result follows by takinga = tk−1 and b = [tk+1, tk+2, . . . , tn−1]in (2) and applying (1)

repeatedly. 

This leaves only the casek = n − 1, for which we use the identity

[t1, . . . , tk−2, tk−1, 1] = [t1, . . . , tk−2, tk−1− 1],

which is obvious as the tail readst_k−1−¹₁=t_k−1− 1.

Remark. One might worry about division by zero. However, if one takes1/0 = ∞and1/∞ = 0, then all the identities in the proof above continue to make sense even with divisions by zero.

Alternatively, it is possible to prove that division by zero does not occur. This can be done by showing in the induction[ti, . . . , tn−1] ≥ 0fori = 1, . . . , n − 1with equality if and only ifi = 1, which requires a few extra case distinctions in the induction step.

Remark. If one does the induction withk = 1ork = n, then the induction step becomes more complicated. Fortunately we did not need to do this thanks to Lemma 1. For example, with k = 1, one needs to prove the implication

[t2− 1, t3, . . . , tn−1] = 0 =⇒ [1, t2, t3, . . . , tn−1] = 0.

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Problemen NAW 5/14 nr. 3 september 2013

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Letb = [t₂, t₃, . . . , tn−1], then this readsb − 1 = 0 ⇒ 1 − 1/b = 0, which is true, but cannot be proved by an identity as in the cases2 ≤k ≤ n − 1, because generallyb − 1 6= 1 − 1/b.

Problem 2013-1/B You are allowed to transform positive integersnin the following way. Write nin base 2. Write plus signs between the bits at will (at most one per position), and then perform the indicated additions of binary numbers. For example,12310= 1111011can get+ signs after the second, third and fifth bits to become11 + 1 + 10 + 11 = 910; or it can get+signs between all the bits to become1 + 1 + 1 + 1 + 0 + 1 + 1 = 610; and so on.

Prove that it is possible to reduce arbitrary positive integers to1in a bounded number of steps.

That is, there is a constantCsuch that for anynthere is a sequence of at mostCtransformations that starts withnand ends at1.

Solution This problem appeared originally in the Fall 2011 edition of the MSRI Emissary. We received solutions from Wouter Cames van Batenburg, Hao Chen, Pieter de Groen, Thijmen Krebs and Guido Senden. The book token goes to Guido Senden. The following solution contains ideas from various solutions.

For a positive integern, letI(n)denote the number of1’s in its binary expansion. For two positive integersa, b, we denote by(ab)2the positive integer obtained by concatenating the binary expansions ofaandb. We remark that ifa, bcan be transformed intox, y, respectively, then(ab)2can be transformed intox + y, by placing a+betweenaandb.

Lemma 3. Every positive integerncan be transformed into every integerxwithI(n) ≤ x ≤

3 2I(n).

Proof. One transformsnintoI(n)by placing+at every position. Omitting a+after a1that has a+before it raises the sum by1, as2 = (1 + 0) + 1and3 = (1 + 1) + 1. This can be done at least j₁

2I(n)k

times, by omitting the plus after every other1. 

Corollary 4. Any positive integernsuch that²₃2^{⌈log2I(n)⌉}≤ I(n) ≤ 2^{⌈log2I(n)⌉}can be transformed into2^{⌈log2I(n)⌉}.

The following proposition, in combination with Corollary 4 now shows that we only need to consider the cases whereI(n) ∈ {4, 5, 8, 9, 10}.

Proposition 5. Suppose that all positive integersnwith8 ≤I(n) ≤ 16can be transformed into 16. Letk ≥ 4, and letnbe a positive integer with2^k−1≤ I(n) ≤ 2^k. Thenncan be transformed into2^k.

Proof. We proceed by induction. For k = 4, this is exactly our assumption. Suppose that the proposition is true fork = i, and letnbe a positive integer with2ⁱ≤ I(n) ≤ 2ⁱ⁺¹. Write n = (ab)2withI(a) =j₁

2I(n)k

andI(b) =l₁

2I(n)m

. Then2ⁱ⁻¹ ≤ I(a) ≤ I(b) ≤ 2ⁱ, soa, bcan both be transformed into2ⁱ, hencencan be transformed into2ⁱ⁺¹, as desired. 

We will now first treat the easiest of the remaining cases, i.e.I(n) ∈ {4, 8, 10}.

Lemma 6. Any positive integernsuch thatI(n) = 4can be transformed into8.

Proof. Writen = (ab)2withaconsisting of the first three bits ofn. Ifastarts with(11)2, then a = 8 − I(b), andbcan be transformed intoI(b)by Lemma 3, soncan be transformed into8. Ifastarts with(10)2, thena = 7 − I(b), andbcan be transformed intoI(b) + 1by Lemma 3, as

I(b) ≥ 2. We deduce thatncan be transformed into8. 

Corollary 7. Any positive integernsuch thatI(n) ∈ {8, 10}can be transformed into16. Proof. Writen = (ab)2, withI(a) = 4. Then bothaandbcan be transformed into8by Lemma 6

and Corollary 4, soncan be transformed into16. 

Finally, we do the caseI(n) ∈ {5, 9}, which requires considering more cases.

Lemma 8. Any positive integernsuch thatI(n) = 5can be transformed into a power of2. Ifn does not start with(11)2, thenncan be transformed into8.

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Proof. First suppose thatnstarts with(11)2. Then writen = (ab)₂, whereaconsists of the first four bits ofn. Ifastarts with(111)2, thena = 16 − I(b), andbcan be transformed intoI(b)by Lemma 3, soncan be transformed into16. Ifastarts with(110)2, thena = 15 − I(b), andb can be transformed intoI(b) + 1by Lemma 3, asI(b) ≥ 2. Hencencan be transformed into16. Otherwisenstarts with(10)2. In this case, writen = (ab)2, wherea = (10)2consists of the first two bits ofn. NowI(b) = 4, which can be transformed into6by Lemma 3, soncan be

transformed into8. _

Lemma 9. Any positive integernsuch thatI(n) = 9can be transformed into16.

Proof. First suppose thatnstarts with(11)2. Writen = (ab)₂withaconsisting of the first three bits ofn. Thena = 4 + I(a) = 13 − I(b). AsI(b) ≥ 6, we have3 ≤ ¹₂I(b). Hence I(b) ≤ I(b) + 3 ≤ ³₂I(b), sobcan be transformed intoI(b) + 3. We deduce thatncan be transformed into16.

Otherwisenstarts with(10)2. Writen = (ab)2in this case, withacontaining the first5ones of n, andbcontaining the last4ones. Then by Lemmas 6 and 8,aandbcan both be transformed

into8, soncan be transformed into16. _

Remark. As Pieter de Groen and Guido Senden remarked, one can prove very quickly that any positive integer can be reduced to1in at most3steps, once one has Corollary 4. For this, note that the remaining cases are those for which2^{⌈log2I(n)⌉−1}< I(n) < ²₃2^{⌈log2I(n)⌉}. The smallest such case isI(n) = 5. Then one can transformninto³₂· 2^{⌈log2I(n)⌉−1}instead by Lemma 3, as

3 2>⁴₃, so

I(n) < ²₃· 2^{⌈log2I(n)⌉}= ⁴₃· 2^{⌈log2I(n)⌉−1}<³₂· 2^{⌈log2I(n)⌉−1}<³₂I(n).

The number ³₂ · 2^{⌈log2I(n)⌉−1}can then be transformed into2asI(n) ≥ 5, which can then be transformed into1, showing that any positive integer can be reduced to1in at most three steps.

Problem 2013-1/C LetRbe a commutative ring with1. Consider the set

S =n

(i, j) ∈ R²:i²=i, j²=j, ij = 0o .

Show that the cardinality ofSis a power of3ifSis finite.

Solution We received solutions from Florian Eisele, Thijmen Krebs and Guido Senden. The book token goes to Florian Eisele. The following solution is partly based on that of Florian Eisele and Thijmen Krebs.

Let us writeS(R)for the setScorresponding to a (commutative unital) ringR. First note that for all idempotentsi ∈ R(i.e. elementsi ∈ Rwithi²=i) we have(i, 0) ∈ S(R). Hence the set of idempotentsI(R)ofRis finite ifSis finite, so we can proceed by induction on the number of idempotentsn(R)ofR. Note that0, 1 ∈ Rboth are idempotents for all ringsR, son(R) ≥ 1, and ifn(R) = 1, thenRis the zero ring, for whichS(R) = {(0, 0)}, hence#S(R) = 3⁰. Ifn(R) = 2, thenI(R) = {0, 1}, thereforeS(R) = {(0, 0), (1, 0), (0, 1)}, and#S(R) = 3¹.

Now letk > 2, and suppose that for all ringsRwithn(R) < k, the setS(R)has cardinality a power of3. LetRbe a ring withn(R) = k. Asn(R) > 2, there exists ani ∈ I(R)withi 6∈ {0, 1}. Letj = 1−i, thenj ∈ I(R), andij = 0. Then the mapϕ : R → R/iR ×R/jR,x 7→ (x +iR, x +jR) is an isomorphism of rings by the Chinese Remainder Theorem. Note thatϕinduces a bijection I(R) → I(R/iR) × I(R/jR), therefore also a bijectionS(R) → S(R/iR) × S(R/jR).

Asjis non-zero, it induces a non-zero idempotent in(R/iR)×(R/jR). It follows thatj+iR 6= 0, so n(R/iR) > 1and similarly (usingi+jR) alson(R/jR) > 1, hencen(R/iR), n(R/jR) < n(R) = k. It follows that#S(R/iR), #S(R/jR)are powers of3, hence so is#S(R) = #S(R/iR)#S(R/jR).