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All use subject to JSTOR Terms and Conditions Solution by John H. Lindsey II, Cambridge, MA. For 0 < x < 1, let f (x) denote the sum of the given power series. Define g by g (0) = 0 and, for x > 0, by g(x) = f

X = j=0 (-_2)Jx2J-1 Now g(1) = 0, because for j in N (-2)j J1(2/j- 1) - =(-I1) jj(-12i - 1)-'+ (-1)j i=1 i=1 i=1 (2-I 1) When 0 < x < 1, we see that g'(x) = -2g(x2), because 0j-1 L(-2)Jx2j-2 j=1

7(2' - 1)-1 = (-2) i=1 co j=1 (-2)

j-1 (x2)2j-1-1H(2i i=1 1)-1

Suppose that f(x) < 0 for some x e (0, 1). Let a = inf{x E (0, 1): f(x) < 0}.

We have 0 < a < 1, f (a) = 0, and f (x) > 0 for 0 < x < a. Hence g(a) = 0. Also,

la 0 = f{xg'(x) + 2xg(x2)}dx =f a xg'(x)dx + a 2xg(x2)dx.

Applying integrationby partsto the firstintegralin the last expression and substituting x = u/112in the second integral yields 0 = ag(a) g(x)dx +

JO g(u)du =

g(u)du < 0,

Ja2 since a2 < a and g(u) > 0 for u < a. The contradiction0 < 0 completes the proof.

Also solved by S. Amghibech (France),K.W.Lau (Hong Kong), R. Richberg(Germany),the GCHQProblem Solving Group(U.K.),and the proposers.

BalancedNeighborhoodSquares 10871 [2001, 372]. Proposed by Donald E. Knuth, Stanford University, Stanford, CA. Say that an n-by-n matrix (aij) is a balanced neighborhood square if the set {(aij, a(i+))(j+ )): 1 < i, j < n} equals {(i, j): 0 < i, j < n} for all eight nonzero choices of (3, E) in {-1, 0, 1}2; subscripts are periodic modulo n. For example, the matrix 3

1 is a balanced neighborhood square. (a) Constructan n-by-n balanced neighborhood squarefor every prime numbern > 4. (b)* Is there any composite numbern > 4 for which an n-by-n balanced neighborhood squareexists? Solution by Gabriel D. Carroll, Harvard University,Cambridge,MA. We answer both parts at once by constructing an n x n balanced neighborhood square whenever n is relatively prime to 6. We work modulo n and assume that all matrix entries are in {0, 1, ...., n - 11. Let aij = (i + 2j)2 + i for every (i, j). It suffices to show that for

161 each (8, E) the map (i, j) -+ (aij, a(i+8)(J+e)) is injective, since this implies that each of the possible n2 values from {0, 1, ... , n - 1}2 occurs exactly once.

Fix 8, e {f-1, 0, 11, not both zero. Let Aij = a(i+6)(j+E) - aij. Now Aij = [(i + 8 + 2j+ 2E)2+ i + 8] - [(i + 2j)2 + i] = 2(i + 2j)(8+ 2E) + (8+ 2E)2+8.

If aij = ai'j, and a(i+8)(j+e) = a(i'+s)(j+e)), then Aij = Airj. Hence 2(i + 2j)(8 + 2E) = 2(i' + 2j')(8 + 26). Since 8 + 26 must lie in {?1, +2, ?3} and is therefore relatively prime to n (just like 2) we can divide both sides by 2(8 + 2E) to obtain i + 2j = i' + 2j'. Using this in aij = apij yields i = i', and hence j = j' as well. Editorial comment.The GCHQ Problem Groupfound a 15x15 squareand conjectured that there is an n-by-n squarewhenever n is sufficiently large.

Both partsalso solved by O. Krafft& M. Schaefer(Germany),S.C. Locke, O.P.Lossers (The Netherlands),R. Stong, L. Zhou, BSI ProblemGroup,andGCHQProblemSolving Group.Part(a) solved by the proposer.

The Fewest3-Cyclesto Generatean Even Permutation 10872 [2001, 469]. Proposed by Bogdan Suceava, Michigan State University, East Lansing, MI. Let a be an even permutationof {1, 2, ... , n}. It is possible to express a as the product of cycles of length 3. Find the minimum number of cycles of length 3 whose product equals o, expressing the answer in terms of the lengths of the cycles in the decomposition of a into disjoint cycles.

Solution by RichardStong, Rice University,Houston, TX.The answeris (n - co(a))/2, where co(a) is the numberof cycles of odd length in the representationof a by disjoint cycles. Let d(a) = (n - co(a))/2.

For the lower bound, we compare the cycle lengths of a with those of a multiplied by the 3-cycle (abc); there are four cases. (1) If a, b, and c are in different cycles of a, then multiplying merges these cycles. (2) If two of {a, b, c} are in one cycle and the thirdis in another,then multiplying replaces those two cycles by two others of the same length. (3) If the elements a, b, and c are in the same cycle of a in that order, then multiplying rearrangesthe cycle without changing its length. (4) If the elements a, b, and c are in the same cycle of a but in the opposite order,then multiplying splits that cycle into three pieces.

In each case, co changes by at most 2. Therefore, Id(a) - d(ar(abc)) I < 1. Since d(identity) = 0, it follows that we need at least d(a) 3-cycles.

For the upperbound, we express a as a productof d(ar) 3-cycles. Since the identity is the only permutationwith d = 0, it suffices to exhibit for every nonidentity permutation a a 3-cycle (abc) such that d(a(abc)) = d(a) - 1. To show this, we produce (abc) that increases the numberof odd cycles by 2.

If a contains a cycle (ala2a3 ... ak) with k > 3, then (abc) = (a3a2al) works, since (ala2a3... ak)(a3a2al) = (a1a4a5... ak)(a2)(a3). If a has no cycle of length at least 3, then since a is even it must contain at least two cycles of length two. If these are (ab) and (cd), then (dcb) works, since (ab)(cd)(dcb) = (abc)(d).

Also solved by S. Amghibech (France),J. Boersema,R. Chapman(U.K.), K. David, J.H.Lindsey,O.P.Lossers (The Netherlands),R. Martin,J.H.Nieto (Venezuela),A. Tissier (France),W. Wardlaw,GCHQProblemSolving Group(UK), Szeged ProblemSolving Group"Fejentalaltuka"(Hungary),andthe proposer. 162

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