Problemen 150

150 150

150 150

150

Pr oblemen

Redactie:

Johan Bosman Gabriele Dalla Torre Ronald van Luijk Lenny Taelman 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 20 euro. 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 100 euro.

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 September 1st, 2011.

Problem A(folklore)

LetSbe an open subset of_R>0that contains arbitrarily small elements. Prove that every positive real number can be written as a sum of finitely many elements ofS.

Problem B(proposed by Gabriele Dalla Torre)

Letnbe a positive integer. Show that every sequence ofnelements of{0, . . . , 9}occurs as a sequence of consecutive digits in the last2ndigits of the decimal representation of some power of2. Also, determine all_{α ∈ R>0}for which the statement still holds if we replace2nby dαne.

Problem C(proposed by Pietro Vertechi)

Letpbe a prime number. Determine the smallest integerdfor which there is a monic polynomial fof degreedwith integer coefficients such thatp^p+1dividesf (n)for all integersn.

Edition 2010-4 We have received correct solutions from R. Kortram, Charles Delorme, Alex Heinis, Rik Bos, Rob van der Waall, Thijmen Krebs, Shai Como, Jos´e Nieto, Anton Schep, Paolo Perfetti, and Moubinool Omarjee.

Problem 2010-4/A Show that there are infinitely many prime numberspfor which there is a positive integernwith

2ⁿ²⁺¹≡ 3ⁿ (modp).

Also, show that there are infinitely many prime numberspfor which there is no suchn.

Solution We received a correct solution from R. Kortram, Charles Delorme, Alex Heinis, Rik Bos, Rob van der Waall and Thijmen Krebs. The book token goes to Rik Bos.

LetPbe the set of primes that divide at least an element of the form2ⁿ²⁺¹− 3ⁿwithna positive integer. The setPis nonempty, because it contains23, and we will prove that it has infinitely many elements.

Suppose, by contradiction, thatPis a nonempty finite set of primes containing23and letnbe Q

p∈P(p − 1). Sincenis greater than1, the integer2ⁿ²⁺¹− 3ⁿhas at least one prime factorq different from2and3. Then, by Fermat’s little theorem we get

2ⁿ²⁺¹− 3ⁿ≡ 1 modq,

contradicting the fact thatqdivides2ⁿ²⁺¹− 3ⁿ. Therefore, the setPcontains infinitely many elements.

151 151

151 151

Problemen NAW 5/12 nr. 2 juni 2011

151

Oplossingen

Letpbe a prime such thatp ≡ 19 mod 24. By quadratic reciprocity neither2nor3is a square modulop. Sincen²+ 1andnhave different parity for every positive integern, the congruence 2ⁿ²⁺¹ ≡ 3ⁿmodphas a square on one side and a non-square on the other one. Hence, it cannot hold andpis not contained inP. By Dirichlet’s theorem on arithmetic progressions there are infinitely many primes congruent to19modulo24. This proves the second part of the problem.

Problem 2010-4/B Letf : R → Rbe a continuous function that has a local minimum or maximum at every point ofR. Show thatfis constant.

Solution We received a correct solution or reference from R. Kortram, Charles Delorme, Shai Como, Jos´e Nieto, Anton Schep, Alex Heinis, Paolo Perfetti, and Thijmen Krebs. The book token goes to Alex Heinis.

Indeed, the following result can be found in several places in the literature.

Proposition. For any functionf : R → R, there are at most countably many_{s ∈ R}for whichf has a local extreme valuesat some point in_R.

Proof. LetSbe the set of values_{s ∈ R}for whichfhas a local maximumsat some point in_R. For eachs ∈ Swe can choose rational numbersa < bsuch that the absolute maximum offon the interval(a, b)equalss. This yields an injective map_{S → Q × Q}sendingsto(a, b), soSis countable. The same holds for the set of values offat local minima.

Now letf be as given in the problem. Iff is not constant, then by the intermediate value theoremftakes uncountably many values, contradicting the proposition.

Problem 2010-4/C Letf : Q × Q → Qbe a function such that for all_{a ∈ Q}the functions x 7→ f (a, x)andx 7→ f (x, a)are polynomial functions from_Qto_Q. Is it true thatfis given by a polynomial in two variables? What if we replaceQbyR?

Solution This problem was solved by Alex Heinis, R.A. Kortram, Jos´e Nieto, Moubinool Omarjee, and Anton Schep. This solution is based on the one by Jos´e Nieto, who wins the book token.

For the casef : Q × Q → Qthe answer is no. The following is a counterexample. Since_Qis countable, there exists an enumerationa1, a2, a3, . . .of_Q. Now define

fn(x) =

n

Y

i=1

(x − ai)

forn ≥ 1, and

f (x, y) =

∞

X

n=1

fn(x)fn(y).

This is well-defined on_{Q × Q}, sincef_n(a_i)equals0forn ≥ i. Furthermore, the specialisation

f (x, ak) =f (ak, x) =

k−1

X

n=1 n

Y

i=1

(ak−ai)(x − ai)

is a polynomial inxof degreek − 1, sofsatisfies the conditions. Iffwere a polynomial inx andy, of a certain total degreed, thenf (x, a_d+2)would be of degree at mostd. Yet we have seen thatf (x, ad+2)has degreed + 1. We conclude thatfis not a polynomial.

Now supposef : R × R → Ris as stated. We will prove that in this casefis indeed given by a polynomial. The elements

pn=x(x − 1) · · · (x − n + 1), n ≥ 0

form a linear basis of the polynomial ring_R[x], so for all_{a ∈ R}we can writef (x, a)as a finite

152 152

152 152

152

Oplossingen

linear combination of the polynomialsp_n. Hence there are functionsc_n: R → Rsuch that the equality

f (x, y) =

∞

X

n=0

cn(y)pn(x) (1)

holds for all_{x, y ∈ R}, while for each_{a ∈ R}, we havecn(a) = 0for all but finitely manyn. The assumption that for eachr ≥ 0, the function

f (r , y) =

r

X

n=0

r ! (r − n)!cn(y)

is a polynomial iny, shows by induction onnthatcn(y)is a polynomial for eachn ≥ 0. We now claim thatcnis identically zero for almost alln ≥ 0. Assume that there are infinitely many n ≥ 0such that the polynomialc_nis non-zero. Each of these polynomials has finitely many zeros, so together they have at most countably many. On the other hand, for all_{a ∈ R}at least one (in fact infinitely many) of these non-zero polynomials has a zero ata, giving uncountably many zeros. This contradiction proves the claim, so (1) shows thatfis itself a polynomial.

R.A. Kortram and Anton Schep refer to the article ‘Some Analogues of Hartog’s Theorem in an Algebraic Setting’ by R.S. Palais, American Journal of Mathematics, Vol. 100. It contains a proof of the following, slightly more general fact: the statement of this problem holds for a fieldkif and only ifkis finite or uncountable.