THE AMERICAN MATHEMATICAL MONTHLY 8-2008

Tạp chí toán học AMM của Mỹ

PROBLEMS AND SOLUTIONS

Edited by Gerald A Edgar, Doug Hensley, Douglas B West

with the collaboration of Paul T Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A Brown, Randall Dougherty, Tam´as Erd´elyi, Zachary Franco, Chris-tian Friesen, Ira M Gessel, Jerrold Grossman, Frederick W Luttmann, Vania Mas-cioni, Frank B Miles, Richard Pfiefer, Cecil C Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull-man, Charles Vanden Eynden, and Fuzhen Zhang

Proposed problems and solutions should be sent in duplicate to theMONTHLY

problems address on the inside front cover Submitted solutions should arrive at that address before December 31, 2008 Additional information, such as gen-eralizations and references, is welcome The problem number and the solver’s name and address should appear on each solution An asterisk (*) after the num-ber of a problem or a part of a problem indicates that no solution is currently available.

PROBLEMS

11376 Proposed by Bogdan M Baishanski, The Ohio State University, Columbus OH.

Given a real number a and a positive integer n, let

an <k≤(a+1)n

1

√

kn − an2.

For which a does the sequence S n (a) converge?

11377 Proposed by Christopher Hillar, Texas A&M University, College Station, TX

and Lionel Levine, Massachusetts Institute of Technology, Cambridge, MA Given a

monic polynomial p of degree n with complex coefficients, let A p be the(n + 1) × (n + 1) matrix with p(−i + j) in position (i, j), and let D p be the determinant of A p

Show that D p depends only on n, and find its value in terms of n.

11378 Proposed by Daniel Troy, Professor Emeritus, Purdue University Calumet,

Hammond, IN Let n be a positive integer, and let U1, , U n be random variables defined by one of the following two processes:

A: Select a permutation of{1, , n} at random, with each permutation of equal probability Then take U k to be the number of k-cycles in the chosen permutation.

B: Repeatedly select an integer at random from{1, , M} with uniform distribu-tion, where M starts at n and at each stage in the process decreases by the value of the last number selected, until the sum of the selected numbers is n Then take U kto

be the number of times the randomly chosen integer took the value k.

Show that the probability distribution of(U1, , U n ) is the same for both processes.

11379 Proposed by Oskar Maria Baksalary, Adam Mickiewicz University, Pozna´n,

Poland, and G¨otz Trenkler, Technische Universit¨at Dortmund, Dortmund, Germany.

Let A be a complex matrix of order n whose square is the zero matrix Show that

R (A + A∗) = R (A) + R (A∗), where R (·) denotes the column space of a matrix

ar-gument

11380 Proposed by Hugh Montgomery, University of Michigan, Ann Arbor, MI, and

Harold S Shapiro, Royal Institute of Technology, Stockholm, Sweden For x ∈ R, let

x

k



= 1

k!

k−1

j=0(x − j) For k ≥ 1, let a k be the numerator and q k the denominator of the rational number−1/3

k



expressed as a reduced fraction with q k > 0.

(a) Show that q kis a power of 3

(b) Show that a k is odd if and only if k is a sum of distinct powers of 4.

11381 Proposed by Jes´us Guillera, Zaragoza, Spain, and Jonathan Sondow, New York,

NY Show that if x is a positive real number, then

e x=∞

n=1

 n



k=0

(kx + 1) (−1) k+1n k

1/n

.

11382 Proposed by Roberto Tauraso, Universit`a di Roma “Tor Vergata”, Rome, Italy.

For k ≥ 1, let H k be the kth harmonic number, defined by H k= k

j=11/j Show that

if p is prime and p > 5, then

p−1



k=1

H2

k

p−1



k=1

H k

k2 (mod p2).

(Two rationals are congruent modulo d if their difference can be expressed as a reduced fraction of the form da /b with b relatively prime to a and d.)

SOLUTIONS

A Group In Which Equations Are Solvable

11231 [2006, 567] Proposed by Christopher J Hillar, Texas A&M University, College

Station, TX Find a non-Abelian group G with the following property: for each positive

integer n, each word W on the alphabet of n + 1 letters A1, , A n and X , each list

a1, , a n of elements of G, and each b in G there exists a unique x in G such that

W (a1, , a n , x) = b (Thus, in particular, ax2ax = b must have a unique solution

x.)

Solution by O P Lossers, Eindhoven University of Technology, Eindhoven, The Netherlands We define such a group onR × R The group operation, denoted ⊕, is defined by

(a, α) ⊕ (b, β) = (a + b, α + e a β).

This operation is associative:

((a, α) ⊕ (b, β)) ⊕ (c, γ ) = ((a + b) + c, (α + e a β) + e a +b γ )

= (a + (b + c), α + e a (β + e b γ ))

= (a, α) ⊕ ((b, β) ⊕ (c, γ )).

Clearly(0, 0) is an identity element Inverses exist, with (a, α)−1 = (−a, −αe −a ) The

operation is not commutative, since in generalα + βe a b

Assume that W contains at least one X We seek a unique solution to

W ((a1, α1), , (a n , α n ), (x, ξ)) = (b, β).

If W has length l, then (b, β) is the result of l − 1 applications of ⊕ Thus b is the sum

of the first coordinates of the pairs in W , and β is the sum of l terms such that the ith

term is the second coordinate of the i th pair in W times the exponential of the sum of the first coordinates of the previous pairs in W

We can thus write (b, β) as (mx + s, σ + μξ), where m is the number of

occur-rences of X in W , and s is an additive combination of a1, , a n Thus x = (b − s)/m,

uniquely determined Furthermore, μ is a sum of exponentials of additive

combina-tions of x and a1, , a n and hence is positive Also,σ is a linear combination of

α1, , α n with coefficients depending on x and a1, , a n Sinceμ is positive, we

must haveξ = (β − σ )/μ.

Also solved by R Bagby, R Chapman (U K.), M Goldenberg & M Kaplan, J Lockhart, D Spellman,

R Stong, C T Stretch (U K.), T Tam, N Vonessen, BSI Problems Group (Germany), Szeged Problems Group

“Fej´ental´atuka” (Hungary), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, and the proposer.

A Special Prime

11235 [2006, 568] Proposed by Lenny Jones and Rachel Keller, Shippensburg

Uni-versity, Shippensburg, PA Find all primes p such that 128 p+ 1 is composite and a divisor of 264 p− 1

Solution by Toni Ernvall and Anne-Maria Ernvall-Hyt¨onen, University of Turku, Fin-land For a fixed prime p, let d be the smallest positive integer such that 2 d ≡ 1

(mod 128p + 1) Suppose that 128p + 1 is composite and is a divisor of 2 64 p− 1

Note that d | 64p and d | φ(128p + 1), where φ is Euler’s function We divide the problem into cases according to whether p divides d.

If p | d, then p | φ(128p + 1) Since p does not divide 128p + 1, there exists a prime q such that q ≡ 1 (mod p) and q | 128p + 1 Write q = rp + 1, so that

128 p + 1 = (rp + 1)(sp + 1) = rsp2+ rp + sp + 1 for some positive integer s, and r sp + r + s = 128 If p = 2, then there are no so-lutions, since 257 is prime Hence p is odd, and r ≡ s ≡ 0 (mod 2) Let r = 2k and s = 2 Now 2kp + k +  = 64, and kp ≤ 31 We may assume that k ≤  By straightforward calculations we notice that the only positive integer solution is p= 3,

k = 1, and  = 9, yielding 128 · 3 + 1 = 385 = 5 · 7 · 11 However, 11 does not divide

264·3− 1, so this solution does not satisfy the original conditions

Assume now that p does not divide d Now d | 64, and 128p + 1 | 264− 1 We compute

264− 1 = (232+ 1)(216+ 1)(28+ 1)(24+ 1)(22+ 1)(2 + 1)

= 641 · 6700417 · 65537 · 257 · 17 · 5 · 3.

Checking through the possibilities gives 128 p + 1 = 65537 · 6700417 Thus p =

3430665851 is the only solution

Also solved by M Avidon, R Chapman (U K.), P Corn, S M Gagola Jr., O P Lossers ((Netherlands)),

A Nakhash, A Stadler (Switzerland), R Stong, M Tetiva (Romania), D B Tyler, BSI Problems Group (Germany), Microsoft Research Problems Group, NSA Problems Group, and the proposers.

An Equality for Parts in Partitions of n

11237 [2006, 655] Proposed by Emeric Deutsch, Polytechnic University, Brooklyn,

NY Prove that the number of 2s occurring in all partitions of n is equal to the number

of singletons occurring in all partitions of n − 1, where a singleton in a partition is a

part occurring once (For example, partitions of 5 yield four 2s: one from(3, 2), two

from(2, 2, 1) and one from (2, 1, 1, 1); partitions of 4 yield four singletons: one from (4), two from (3, 1) and one from (2, 1, 1).)

Solution I by John H Smith, Needham, MA Let p (n) denote the number of partitions

of n, with p (0) = 1 and p(n) = 0 for n < 0 The number of partitions of n with at

least i occurrences of j is p (n − i j) Hence the total number of 2s in all partitions of

n is ∞

i=1 p (n − 2i), since a partition with exactly k occurrences of 2 is counted once

in each of the first k terms.

On the other hand, a partition of n with a singleton j is formed by appending j to

a partition of n − j that has no j Since p(n − 2 j) partitions of n − j contain at least one j , the number of partitions of n with a singleton j is p (n − j) − p(n − 2 j) Thus

the total number of singletons in partitions of n− 1 is ∞

j=1(p(n − 1 − j) − p(n −

1− 2 j)) This sum equals

(p(n − 2) − p(n − 3)) + (p(n − 3) − p(n − 5)) + (p(n − 4) − p(n − 7)) + · · ·

Terms p (n − r) occur positively when r is even and with both signs when r is odd.

Cancellation yields ∞

i=1p (n − 2i), which is the formula obtained for the total number

of 2s

Solution II by Albert Stadler, D¨ubendorf, Switzerland Euler obtained the generating

function for partitions:

P (x) =∞

n=1

p (n)x n =∞

n=1

1

(1 − x n ) .

The number of partitions of n with exactly k occurrences of 2 is the coefficient of x nin

x 2k (1 − x2)P(x) Hence the total number of 2s over all partitions of n is the coefficient

of x nin each expression below:

∞



k=1

kx 2k (1 − x2)P(x) = x2

(1 − x2)2(1 − x2)P(x) = x2

(1 − x2) P (x).

On the other hand, the number of partitions of n − 1 containing the singleton k is the coefficient of x n−1 in x k (1 − x k )P(x) Hence the total number of singletons over all

partitions of n − 1 is the coefficient of x n in x ∞

k=1x k (1 − x k )P(x) Note that x

∞



k=1

x k (1 − x k )P(x) = x x

1− x −

x2

1− x2

P (x) = x2

(1 − x2) P (x).

Thus the coefficients are equal, as claimed

Also solved by M Avidon, D Beckwith, N Caro (Brazil), R Chapman (U K.), J Dalbac, P P D´alyay (Hun-gary), Y Dumont (France), G Keselman, M Kidwell, S C Locke, O P Lossers ((Netherlands)), P Massaro (Italy), R McCoart, M D Meyerson, A Nijenhuis, I Novakovic & A Milan (Serbia), R Pratt, R Stong,

R Tauraso (Italy), J Taylor-Goodman, M Tetiva (Romania), W Watkins, BSI Problems Group (Germany), Szeged Problems Group “Fej´ental´atuka” (Hungary), GCHQ Problem Solving Group (U K.), Microsoft Re-search Problems Group, NSA Problems Group, and the proposers.

Sequences Built from the Golden Ratio

11238 [2006, 655] Proposed by Aviezri S Fraenkel, Weizmann Institute of Science,

Rohovot, Israel Let n be(n + 1)φ φ − nφ φ, where φ = (1 +√5)/2 and

x denotes the integer part of x Prove that the following hold for every positive integer n:

(a) n is either 2 or 3;

(b)

(n + 1)φ2 φ − n φ2 φ = 2 n− 1;

(c)nφ +n φ2 = n φ2 φ ;

(d)

n φ2 φ =nφ φ2 + 1

Solution by Reiner Martin, Teddington, U.K Since n − 1 < (nφ − 1)/φ ≤ nφ /φ <

n, we have nφ /φ = n − 1 With φ = 1 + 1/φ, this implies

nφ φ = nφ (1 + 1/φ) = nφ + n − 1.

Thus n = (n + 1)φ − nφ + 1, and so (a) follows since 1 < φ < 2.

Note that nφ φ = nφ + nφ /φ ≤ nφ + n = n(φ + 1) = n φ2 Thus

nφ ≤ n φ2 /φ ≤ nφ, and so n φ2 /φ = nφ Thus n φ2 φ =n φ2 +

n φ2 /φ =n φ2 + nφ, which is claim (c).

Moreover, it follows that

n φ2 φ = 2 nφ + n Applying this to n + 1 and to n

and invoking n = (n + 1)φ − nφ + 1 yields (b).

Finally, usingnφ φ = nφ + n − 1, (d) follows from

n φ2 φ = 2 nφ + n = nφ + nφ φ + 1

= nφ (1 + φ) + 1 =nφ φ2 + 1.

Also solved by M R Avidon, D Beckwith, N Caro & O L´opez (Brazil), R Chapman (U K.), J Christopher,

P P D´alyay (Hungary), M Goldenberg & M Kaplan, G Keselman, R Lampe, J H Lindsey II, O P Lossers ((Netherlands)), A Nijenhuis, S Northshield, J Rebholz, K Schilling, A Stadler (Switzerland), A Stenger,

R Stong, BSI Problems Group (Germany), Con Amore Problem Group (Denmark), GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, and the proposer.

Perfect Parity Patterns

11243 [2006, 759] Proposed by Donald Knuth, Stanford University, Stanford, CA An

m × n matrix of 0s and 1s is a parity pattern if every 0 is adjacent (horizontally or

vertically) to an even number of 1s and every 1 is adjacent to an odd number of 1s It

is perfect if no row or column is entirely zero Thus,

⎛

⎝10 10

1 1

⎞

⎠ ,

⎛

⎜

⎝

⎞

⎟

⎠ ,

⎛

⎝01 11 00 11 01

⎞

⎠ ,

⎛

⎜

⎜

⎝

⎞

⎟

⎟

⎠

are parity patterns of sizes 3× 2, 4 × 4, 3 × 5, and 5 × 5, respectively; only the 4 × 4 and 5× 5 patterns are perfect

(a) Determine the number c (n) of perfect parity patterns that have exactly n columns.

(b) Alone among these examples, the 5× 5 parity pattern is invariant under rotation

by 90 degrees and under reflection across its central column Thus it has eightfold

symmetry Prove that a perfect n × n parity pattern with eightfold symmetry exists for all n of the form n = 3 · 2k − 1 with k ≥ 1.

Solution of part (a) by Robin Chapman, University of Bristol, Bristol, U K Let c (n)

be the number of perfect parity patterns with n columns To avoid a trivial addition of

1 to the formula, we require at least one row We prove that

c (n) = 

d |(n+1)

μ n+ 1 d

2d−1,

whereμ denotes the M¨obius function.

Let M be an m × n matrix, with rows v1, , v m Let 0 denote the all-zero vector

of length n, and set v0 = v m+1 = 0 In binary arithmetic, the condition that M is a

parity pattern is that the sum of entry(i, j) and its neighboring entries in M is 0 (that

is, even) Equivalently,

for 1≤ i ≤ m, where T n is the n × n binary matrix whose (i, j) entry is 1 if and only

if|i − j| ≤ 1.

Note that(∗) is a recurrence that determines v i+1 fromv i−1 andv i Also, it is re-versible: fromv i andv i+1, it determinesv i−1 Fixingv0 = 0, any n-vector v1 deter-mines a doubly infinite sequence(v i )∞

i=−∞of vectors via(∗) Since F n

2is finite, some pair(v i , v i+1) must repeat as (v i +p , v i +1+p ) for some p Now (∗) implies that v i = v i +p

for all i , and hence the sequence is periodic.

Call a parity pattern semi-perfect if it has no all-zero row (it may have all-zero

columns) A semi-perfect parity pattern is determined by its first row v1, which is

nonzero Since the sequence is periodic, there is a first positive integer q such that v q =

0 The parity pattern with rowsv1, , v q−1is the only semi-perfect parity pattern with

first rowv1 As there are 2n − 1 nonzero vectors of length n, there are 2 n− 1

semi-perfect parity patterns with n columns.

We now claim that in the sequence produced byv0andv1with q defined as above,

v i = 0 if and only if q | i Suppose v k = 0 From v k+1 = v k T n + v k−1, we obtain

v k+1 = v k−1 Applying the same computation to v k + j shows inductively thatv k + j =

v k − j for all j Thus the sequence v j  is symmetric about any k for which v k= 0 Since

v0= 0 and v q = 0, it follows that v mq = 0 for all m Since v j

our symmetry observation,v j

quence, the number of rows in a parity pattern with first rowv1must be one less than a

multiple of q Let M k (v1) denote the parity pattern with kq − 1 rows and first row v1;

note that only k= 1 yields a semi-perfect parity pattern

Now consider an m × n parity pattern M that is semi-perfect but not perfect; M has

an all-zero column Let the first all-zero column be in position r If r = 1 then by the

definition of a parity pattern, the other columns are forced to be zero, so r > 1.

The matrix N formed by the first r− 1 columns is a parity pattern with no all-zero

column Its transpose N T is a semi-perfect parity pattern Hence N T = M1(w), where

w is the transpose of the first column of M Now M T = M k (w) for some integer k

with k ≥ 2 By the preceding paragraph, n = kr − 1 As M is semi-perfect, M T

has

no zero column Since every row of M T either equals a row of N T or is zero, it follows

from the lack of all-zero columns in M T that N T has no all-zero column Hence N is

perfect

On the other hand, from a perfect parity pattern N with r− 1 columns and first columnw, we can reverse this process to obtain a semi-perfect parity pattern M with

kr − 1 columns: let M = M k (w) T , where N = M1(w) T Since N has no zero rows, neither does M Thus M is semi-perfect and its first all-zero column is in position r Counting semi-perfect parity patterns with n columns in groups according to the associated perfect parity pattern N gives

r |(n+1),r>1 c (r − 1) = 2 n − 1 Setting j =

n + 1, a1 = 1, and a r = c(r − 1) for r > 1, this becomes r | j a r = 2j−1 By M¨obius

inversion, a j = d | j μ( j/d)2 d−1 This is equivalent to the claimed result for c(n)

since c (n) = a n+1.

Solution of part (b) by Reiner Martin, Teddington, U K We recursively construct a

perfect parity pattern P kwith eightfold symmetry and 3· 2k− 1 rows

Let P1be the 5× 5 parity pattern in the problem statement With a i , jdenoting entry

(i, j) in P k−1, define entry(r, s) in P k to be b r ,s, where

b 2i ,2 j = a i , j ,

b 2i +1,2 j = a i , j + a i +1, j (mod 2),

b 2i ,2 j+1 = a i , j + a i , j+1 (mod 2), and

b 2i+1,2 j+1 = 0, where a i , j is understood to be 0 if i or j equals 0 or 3· 2k−1 It is straightforward to

verify the parity condition for each type of entry; hence P kis indeed a parity pattern

Also, eightfold symmetry is preserved by the construction A zero row in P k would

imply two adjacent identical rows in P k−1, which by induction does not occur Thus

P kis indeed perfect

Editorial comment Parity patterns arise in the study of a popular puzzle called

“Lights Out”, invented in the 1980s by Dario Uri and available as a Java applet at http://www.whitman.edu/mathematics/lights_out/ Klaus Supner called such

pat-terns “even-parity covers” of the m × n grid; see Math Intell 11 No 2 (Spring 1989) 49–53, and Theor Comp Sci 230 (2000) 49–73.

Also solved by D Beckwith, S M Gagola, D Serre (France), Microsoft Research Problems Group, and the proposer Part (a) also solved by J H Lindsey II.

Double Integral with Log and Sinh

11275 [2007, 164] Proposed by Michael S Becker, University of South Carolina at

Sumter, Sumter, SC Find

 ∞

y=0

 ∞

x =y

(x − y)2log((x + y)/(x − y))

x y sinh (x + y) d x d y . Solution I by David Beckwith, Sag Harbor, NY Let I denote the required integral.

Introduce the change of variables x = (z + t)/2, y = (z − t)/2 which gives

I = 2

 ∞

z=0

1

sinh z

 z

t=0

t2log(z/t)

z2− t2 dt d z

In the inner integral, set t = zw to obtain the product of two known integrals,

I = 2

 ∞

0

z d z

sinh z

 1

0

1− w2

logw dw = 2 π2

4

π2

8 − 1

= π2(π2− 8)

Solution II by Hongwei Chen, Christopher Newport University, Newport News, VA.

Let I denote the required integral Substitute x = uy and interchange the order of

integration to obtain

I =

 ∞

u=1

(u − 1)2log((u + 1)/(u − 1))

u

∞

y=0

y

sinh(y(u + 1)) d y

du

Let I u denote the inner integral in the previous formula for I Substituting t = e −(u+1)y

gives

I u = −2

(1 + u)2

 1

0

ln t

t2− 1dt =

−2

(1 + u)2

 1

t=0

∞



k=0

t 2k ln t dt

(1 + u)2

∞



k=0

1

(2k + 1)2 = π2

4(1 + u)2.

So I takes the form

I = π2 4

 ∞

1

(u − 1)2log((u + 1)/(u − 1))

Finally, substitute s = (u − 1)/(u + 1):

I = −π2

2

 1 0

s2ln s

1− s2 ds= −π2

2

 1 0

ln s

1− s2 − ln s

ds = π2(π2− 8)

Editorial comment Many other interesting substitutions were submitted.

Also solved by R Bagby, D H Bailey (USA) & J M Borwein (Canada), R Chapman (U K.), K Dale (Norway), B E Davis, P De (Ireland), N Eklund, C Fleming, W Fosheng (China), M L Glasser, M R Gopal, R Govindaraj & R Mythili & G Anupama & G Sudharsan (India), J Grivaux (France), J A Grzesik,

E A Herman, G Keselman, O Kouba (Syria), G Lamb, D Lovit, K McInturff, O Pad´e (Israel), P Perfetti (Italy), T.-L & V R˘adulescu (Romania), O G Ruehr, H.-Z Seiffert (Germany), A Stadler (Switzerland), R Stong, R Tauraso (Italy), E I Verriest, M Vowe (Switzerland), FAU Problem Solving Group, GCHQ Problem Solving Group (U K.), Microsoft Research Problems Group, University of Sharjah Problem Solving Group (United Arab Emirates), and the proposer.

A Matrix Inequality

11296 [2007, 252] Proposed by Said Amghibech, Quebec, Canada Let A be a real

symmetric positive definite matrix, with entries a i , j, 1≤ i, j ≤ n Let A k be the k × k matrix in the upper left corner of A Show that

n



k=1

(det A k )1/k≤ 1+ 1

n

n

Tr(A)

where Tr(A) denotes the trace of A.

Solution by Joanna Chachulska and Wojciech Matysiak, Polytechnica Warszawska, Warsaw, Poland Hadamard’s inequality says det A k ≤ a1,1 a2,2 · · · a k ,k Carleman’s

inequality says

n



k=1



x1x2· · · x k

1/k≤ 1+1

n

n n

k=1

x k

for x k≥ 0 Our result follows

Note: Carleman’s inequality is usually stated in its limit form, with infinite sums and

the number e replacing the term (1 + 1/n) nby monotonicity However, standard proofs

of the inequality prove the finite form stated above For example, see pp 146–152 of

G P´olya, Mathematics and Plausible Reasoning, Princeton Univ Press, Princeton, NJ,

1954

Also solved by D Chakerian, R Chapman (U K.), J.-P Grivaux (France), E A Herman, C J Hillar, F Holland (Ireland), D Jesperson, O P Lossers (Netherlands), M Omarjee, H.-J Seiffert (Germany), M Tetiva (Romania), NSA Problems Group, and the proposer.

definition of a parity pattern, the other columns are forced to be zero, so r > 1.

The matrix... only the × and 5× patterns are perfect

(a) Determine the number c (n) of perfect parity patterns that have exactly n columns.

(b) Alone among these examples, the. .. infinite sums and

the number e replacing the term (1 + 1/n) nby monotonicity However, standard proofs

of the inequality prove the finite form stated above