Solution 4 It shouldbenotedthatthisproblemwasseverly underestimatedby theproblemscom-mittee,whosesolutionsall requiredtacitadditionalhypothesesandwouldnot have scoredmorethan4 outof 10. In fact, theproblemis harderthanProblem6.

Let A and B be any pair of friends. Togetherwith their mutual friend they form a ‘friendshiptriangle’. Noneof thesethreefriendshipscanoccurin anothertrianglewithout violating uniquenessof mutualfriendship.The total numberof friendshipsequals3t � wherethenumberof trianglesis t �EachpersonX hastwo friendspertriangle,andthereforeanevennumberof friends.

Now takeany triangleABC � ThepeopleotherthanA � B andC canbedivided into four mutuallyexclusive sets:otherfriendsof A � B andC; and‘strangers’.Let thesesetscontainrespectively a � b �c ands people. Now the mutual friend of any pair of other friendsof A andB mustbe a stranger,otherwiseuniquenessof mutualfriendshipis violated.(To seethis, it is easiestto referto a diagram.)Ontheotherhand,everystrangercanbeidentifiedin auniquewayfrom thepairof hismutualfriendswith A andB respectively. Thereareab suchpairs.Therefore

s � ab � bc � ca �wherethelasttwo equationsfollow from symmetry.

If s � 0 � two of a � b andc mustalsobezero.Thepersoncorrespondingto thenonzerovalueis thena friendof everyoneelse.This will turnout to betheonly casepossible.

If s�� 0 � thena � b � c � which impliesthateverybodyhasthesamenumberof friendsd � We can

countthetotal numbern of peoplein termsof d : otherfriendsof A � B andC tally � d � 2� each,andstrangers� d � 2� 2 � Thisgives

n � 3 � 3 � d � 2���� d � 2� 2 � d2 � d � 1 �Thelastandmostdifficult partof theproof is to show thatfor d 2 � thiscountfor n is incompatible

with theotherconditions,which eliminatesthecases�� 0 � therebyproving the theorem.Remember

thatd mustbeeven,andthat thetotal numberof friendshipsis 3t � Notethat therefore3t � 12nd � We

startwith thecased � 4 � Thisgivesn � 13 and 12nd � 26� which is not amultiple of 3.

For d not of theform 6 j � 2 thesameargumentgivesno contradiction.We needto find a general-izationof a trianglewhich canbecountedin two ways,suchthat the two countscannotpossiblybeequal.A suitableobjectto countis a ‘friendshipcircle’ A1A2 ����� Ak � in which eachpersonis a friendof thetwo peopleadjacentto him, with A1 adjacentto Ak � andwhereit is not requiredthatall Ai bedistinct.A friendshipcircle reducesto a trianglewhenk � 3 � Let mk bethenumberof suchcircles.

Now A1 canbeany of n people,andeachAi from A2 to Ak � 1 any of thed friendsof hispredecessor,giving ndk � 2 possibilities.At thatpoint therearetwo cases:

1. If Ak � 1 � A1, thenA1A2 ��� � Ak � 2 is a friendshipcircle,which canhappenin mk � 2 ways.Ak canbeany of thed friendsof A1 � whichgivesusdmk � 2 possibilities.

2. If Ak � 1�� A1, thenthereis no furtherfreedom— Ak mustbetheuniquefriend of Ak � 1 andA1 �

Thisgivesusndm � 2 � mk � 2 possibilities.

We thereforeget:

mk � dmk � 2 � ndm � 2 � mk � 2 ��� d � 1� mk � 2 � ndm � 2 �Sincen � d � d � 1��� 1 � it followsthatmk � 1 mod � d � 1� whenk 2 �

On the other hand, whenever k is a prime, the friendship circles A1A2 ����� Ak � A2 ��� � AkA1 � ����� �AkA1 ����� Ak � 1 � are all different. (To seethis, imaginepeoplemoving up a fixed numberof placesarounda circular table.) Thereforefriendshipcirclescomein mutually exclusive classesof k itemseach,which impliesmk � 0 modk �

2

To completetheproof,notethatd � 1 is oddandmustthereforehaveaprimefactork 2 wheneverd � 4 � This leadsto mk � 1 modk � which givestherequiredcontradiction.

Solution 6 Therequiredresultis aspecialcaseof thefollowing:

n squaresof total areaQ canbe packedinto a rectangleof area2Q provided that the largestof thesquarescanfit into therectangle.

We prove this resultby induction.Thecasen � 1 is trivial.In the proof, we shall repeatedlyusetheeasilyproved fact that if two squareshave total areas2 �

their sidescannotsumto morethan � 2s � which is attainedwhenthesquaresareequal.In thegeneralcase,denotethesidesof therectangleby a andb with a � � 2Q � b � thek-th largest

squareby Sk andits sideby sk �If s1 a � 2 � we cancut therectangleinto two pieces:oneof dimensions1 � a containingS1 � and

the otherof dimension � b � s1 � � a into which theothern � 1 squaresof total areaQ � s21 areto be

packed.Sincea � b � s1 ��� ab � 2s21 � 2 � Q � s2

1 � � this is possibleby theinductionhypothesisprovidedthats1 � s2 � b � This is alwaysthecasebecauses2

1 � s22 � Q whereasb � � 2Q �

We areleft with thecases1 � a � 2 � It is thenpossibleto packS1 � S2 � S3 � S4 into the rectangle,oneinto eachcorner, sothatS1 andS2 adjoinoneof theshortersidesof therectangle,andS1 andS3 adjoinoneof thelongersides.Wenow arguethatit is possibleto fit S5 alongthatsidebetweenthem,i.e. thats1 � s3 � s5 � b �

Clearlythehardestcaseoccurswhens2 � s3 ands4 � s5 � Thens3 � s5 ��� 2 � s23 � s2

5 ����� Q � s21

sincethefive squareshave total areanot exceedingQ � But thelargestpossiblevalueof s1 � � Q � s21

is � 2Q � b �Wenow chooseasetof squares,startingwith S1 andS2 andcontinuingwith S5 andsmallersquares,

sothattheir totalareaQ1 is at leastas1 � 2 � but removing any squarefrom thesetreducesthetotalareato lessthan as1 � 2 � The idea is that c � 2Q1 � a shouldexceeds1 by as little as possible. In fact,c � s1 � s2

5 ��� a � 2��� s25 � s1 � s5 �

Cut therectangleinto two pieces,oneof dimensiona � c wherec � 2Q1 � a � into which this setofsquarescanbepackedby theinductionhypothesissinceby constructions1 � 2Q1 � a; andtheotherofdimension2 � Q � Q1 ��� a � into whichtheremainingsquarescanbepackedby theinductionhypothesis,providedthats3 � b � c �

But we showedearlierthats1 � s3 � s5 � b � hences3 � b ��� s1 � s5 ��� b � c asrequired.