Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at http://www.jstor.org/page/ info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact support@jstor.org.

Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The American Mathematical Monthly. http://www.jstor.org

This content downloaded from 129.199.129.11 on Fri, 15 Jan 2016 20:57:10 UTC

All use subject to JSTOR Terms and Conditions j shows that

E -j=0-(N1)J Also solved by R. Chapman(U.K.), GCHQProblemSolving Group,andthe proposer.

F(m, j) is independent of m. When m = 1, the sum reduces to ),which indeed equals 1.

10873 [2001, 469]. Proposed by B. J. Venkatachala,Indian Institute of Science, Bangalore, India. Find every pair (m, n) of nonnegative integers that satisfies the equation 3m = 2n2 ? 1.

Solution by Doyle Henderson, Omaha, NE. The only possibilities for m are 0, 1, 2, 3, and 5. Simple calculations then show the solution set for (m, n) to be {(0, 0), (1, 1), (2, 2), (5, 11)1}.

Suppose there is a solution with m even and at least 4. Let m = 2a and x = 3a We have x2 - 2n2 = 1. Since this is a Pell equation, it follows that x = Pj for some nonnegative integer j, where Po = 1, P1 = 3, and Pk = 2Pk-1 + Pk-2 for k > 2. Since a > 2, it follows that 9 divides Pp. Since Pk+12= 9 11 - 140Pk+1 + 5741Pk, and P5 is the only element of {Po, ... , P11}that is divisible by 9, it follows inductively that j = 5 mod 12. Induction then shows that 11 divides Pj, which is a contradiction.

Now suppose there is a solution with m odd and at least 7. Let m = 2b + 1 and y = 3b and z = 2n. We have Z2 - 6y2 = -2. This is anotherPell equation, so y = Qj for some nonnegative integer j, where Qo = 1, Q1 = 2, Q2k = 4Q2k-1 ? Q2k-2, and Q2k+1 = 2Q2k + Q2k-1 for k > 1. Since b > 3, we find that 27 divides Qj. Since Q2k+9 = 27 .17 19Q2k+1 + 1960Q2k, and Q2k+10 = 27 .17 19Q2k+2 + 3920Q2k+1, and Q8 is the only element of {Qo, Q1 ... Q8} divisible by 27, induction shows that j - 8 mod9. It thenfollowsinductivelythat17 dividesQj, whichis a contradiction. EditorialcommentA.ndrzejMakowskinotedthatin Mathesis(2) 4 (1894) 169-170,E. Fauquemberguperovedthat 3i y2 onlywhent E {0, 1, 4} (see L. E. Dickson, History of the Theory of NumEb- ers, v-ol. 2, Chelsea, New York, 1966, p. 694). This yieldsthe solution.

Manysolversobtainedthe resultdirectlyfromW. Ljunggren,Some theoremson indeterminateequations of the form xn-1 = yq, Norsk Mat. Tidsskr.25 (1943) 17-20. Also solved by S. Amghibech (France),M. Bowron, R. Chapman(U.K.), F. Coghlan (U.K.), S. M. Gagola Jr., GCHQ Problems Group (U.K.), J. Grivaux(France), C. J. Hillar, O. P. Lossers (Netherlands),F. Luca & S. Hernandez(Mexico), R. Richberg(Germany),R. Stong, andthe proposer.

10875 [2001, 469]. Proposed by Donald E. Knuth, Stanford University,Stanford, CA. A polyomino spans a squareif it can be embedded in the squaretouching all four sides. There are 6 pentominoes that span a 3-by-3 square:

Solutionby RichardStong,Rice UniversityH,ouston,T + 2n-2 3n2-2n+8+ E, where E = 0 if n is even and E = 22-1(T/2T(. hIe ans w-eriifs n(is-2o1dd. We first cou8nt the diagrams of polyominoes that span the sq(una-1r)e/2a)nd+th einn, apply Burnside's lemma to count the equivalence classes underrotation and reflection.

We view each suchdiagramas a subgraphof a gridrepresentingthe squares.The graphof a polyominohas a vertexfor each cell in the polyomino,withtwo vertices adjacentwhenthecorrespondingcells sharea side. By thedefinitionof a polyomino, the graphis connected.We place the n-by-n squarein the firstquadrantand name each possible cell (and the correspondingvertex)using an orderedpair (i, j) with i, j {0, ... , n - 1}, indexed from the lower left.

The graphof eachpolynominohas 2n - 1 vertices.We claimalso thatit is a tree with n - 1 horizontal edges and n - 1 vertical edges. It must have at least n - 1 edges of eachorientationin orderto havepathsconnectingtheextremerowsandtheextreme columns.Froma givenvertex,a pathcanaddonly one newrowor columnwitheach newvertex.Since2n - 2 additionarlowsandcolumnsmustbe addedandonly2n - 2 additionalverticescanbe added,the additionavlerticescannotcompletea cycle, and thegraphis a tree.

The requiremenotf monotonicityimplies that all the horizontaledges lie in the uniquepathjoiningtheextremehorizontaledges,thatthispathneverreversesvertical direction,and thatthe remainingedges form a segmentstraightup from a highest vertexon the pathanda segmentstraightdownfroma lowestvertexon it. Described symmetricallys,uchgraphsarethose consistingof a monotonelatticepathP within thesquareandsegmentsto thefoursidesfromtheendsof P. Furthermores,uchgraphs do correspondto desiredpolyominoes,sincetheygrowpathsfromone vertexto span the squarewith 2n - 1 cells.

We countthese graphsbeforeapplyingthe geometricsymmetries.We obtainthe graphsfromup-rightlatticepathsextendingfrom (0, 0) to (n - 1, n - 1). Most of the latticepathsgenerateeightgraphs,andmostgraphsariseonce,buttherearesome degeneratecases.

A latticepathis itself one of the desiredgraphs.To generateadditionalgraphs, considera latticepathwith at least two changesof direction,and supposethatthe secondchangeof directionoccursatthepoint(k, 1).Replacethefirststraightsegment of thepath(in thebottomrowor leftmostcolumn)so thatthe graphnow includesthe straightsegmentsto (k, 1)from(0, 1)and(k, 0).

If the graphis generatedin this way andits centralmonotonepathP is nontrivial, thenthelatticepaththatgeneratesit canbe retrievedT.hepoint(k, 1)is thefirstpoint of P. If P continueswitha verticaledge,thena verticalsegmentwasmovedto obtain it fromthelatticepath(otherwiset,hehorizontaslegmentwouldhavemovedupfarther than1).Similarlyi,f P startswitha horizontaledge,thenthelatticepathbeganwitha horizontalsegment.Wepostponethecase when P is a singlepoint.

The analogousoperationcanbe performedat the end of the latticepath.Also, the pathandresultinggraphcan be reflectedthroughthe line y = (n - 1)/2. Fora path withmorethantwo changesof direction,we thusobtaineightgraphs.Fora pathwith exactlythreechangesof direction,applyingthemodificationatbothendsyieldsa + shapedgraphin whichP is a singlevertex.Thereare(n - 2)2choicesforthelocation of P in thiscase, andeachsuchgrapharisesfromfourof thepaths.

Therearetwo pathswithonly one changeof direction(at (n - 1, 0) or (0, n - 1)), and these generate the four graphs that are these paths and their reflections, not 16 graphs. For a lattice path with exactly two changes of direction, we can still apply the eight operations, but in some cases this generates only four graphs, each arising twice. We include these in the eight-per-pathcount, because we adjust for the overcount by subtraction.

The graphs generated from lattice paths with exactly two changes of direction are Z-shaped (a lattice path or its reflection) or T-shaped (one border and a straightpath perpendicularto it). Each such grapharises twice, and there are 8(n - 2) such graphs. 244

@ THEMATHEMATICAALSSOCIATIONOFAMERICA [Monthly 110 Thus the total numberof graphs is 8(2n-2) - 12 - 3(n - 2)2 - 8(n - 2)

To count the polyominoes, we apply Burnside's lemma to count the equivalence classes of these graphs under the symmetries of the square. We sum the numbers of graphsinvariantundereach symmetry and then divide by 8, the numberof symmetries. All the graphs are invariantunder the identity operator.

A graphinvariantunderreflection throughthe diagonal from (0, n - 1) to (n - 1, 0) arises from a lattice path determinedfrom its firstn - 1 steps by symmetry,modified at both ends or neither,and not reflected throughy = (n - 1)/2. The n graphsthatare + shapes with centers on this diagonal (or lattice pathswith one change of direction) arise twice in this count. Hence there are 2n-1 - n graphsinvariantunderthis reflection. The numberinvariantunder reflection through the other diagonal is the same.

To be invariantunder horizontal or vertical reflection, a graph must be a + -shape or T-shape with vertices along the axis of reflection. Hence such graphs require n to be odd, and then for each axis there are n such graphs.

All graphs invariantunder nontrivial rotations must contain the center point Q at ((n - 1)/2, (n - 1)/2), so these occur only when n is odd. The only graph invariant undereither of the 900 rotations is the + -shape with center at Q.

A graph invariantunder 1800 rotation arises from a lattice path from (0, 0) to Q by modifying both ends or neither and possibly reflecting. Thus each such lattice path generates four such graphs, except that the two lattice paths with one change of direction together generate only five such graphs (four Z-shapes and the central + -shape). Hence there 4((n-l/2) - 3 graphs invariantunder this rotation.

For even n, we add the 2" - 2n diagonal invariants and divide by 8 to obtain 2n-2 polyominoes. For odd n, adding the terms for invariance (2n-2 3n2-2n+8 under all symmetries yields (n-2) + 2n-2 3n2-4n+98 2 ((n-1)/2 Also solved by D. Beckwith, J. H. Lindsey II, GCHQProblemSolving Group,andthe proposer.

10877 [2001, 469]. Proposed by Emeric Deutsch, Polytechnic University, Brooklyn, NY An L-tile is a 2-by-2 square with the upper right 1-by-1 subsquare removed; no rotations are allowed. Let an be the numberof tilings of a 4-by-n rectangle using tiles that are either 1-by-i squaresor L-tiles. Find a closed form for the generating function 1 + aix + a2x2 + a3x3 .

Solution by Daniele Donini, Bertinoro, Italy. We use a system of recurrencesto prove that the generating function is (1 - x2 - x3)/(1 - x - 5x2 - 4x3 + x5).

For n > 1, let bn, Cn, anddnbe the numberof ways of tiling a 4-by-n rectangle where in the first column we remove the bottom square, the bottom two squares, and the top two squares, respectively. Observe that ao = 1 (counting the empty tiling) and that al = bl = cl = dl = 1, since with only one column the only tiling uses all squares.

For the 4-by-n rectangle with n > 2, there are an-1 tilings where no L-tile meets the first column (squares fill that column). There are an-2 tilings where two L-tiles intersect the first column (the second column has two squares) . There are bn-1 + cn-1 + dn-1 tilings where one L-tile intersects the first column (depending on how high it is). sis,Tbhnu=s foarn-n1 >+ 2cnw-1e +hadvne-1a,n c=, a=n-a1n-+1 a+n-2 ? bann-1d +dnc=,1 an+-1 d+n-1b.n-B1.y Lsiemttiilnagr aan1_al=ybo = co = do = 0 extends the validity of thdenr-e1c,urrencesto n > 1.

Let A, B, C, D be the generating functions for these four sequences. Multiplying

PROBLEMASNDSOLUTIONS 245