PerfectSquares,Cubes,FourthPowersandMore
Manyproblemsinnumbertheoryinvolveexpressionswithperfectsquares,perfectcubes
orperfectfourthpowers.Thereareproblemsthatrequireustofindintegersolutionsto
equationswithunknownexponents.BelowIwilllistveralcommontypesofproblems
andstrategiestotacklethem.Ofcour,thelistisnotexhaustive.
1.Findingintegersolutionstoanequationinvolvingperfectsquares,cubesor
fourthpowers.Manyoftheproblemscanbetackledbyconsideringthepos-
sibleresidues,moduloasuitablenumber.Forperfectsquares,tryingconsidering
modulo4,8,16,3,5,7,11or13.Forperfectcubes,itmaybeufultotake
modulo7,9or,ocassionally,modulo13.Forperfectfourthpowers,youmight
wishtoconsidermodulo5or16.Ofcour,thearejustsuggestions,andit
reallydependsontheproblem.
Forequationsthatinvolveonlyperfectsquaresintwovariables,itmaybeufulto
considertheequationasaquadraticinonevariable,anduthefactthatdiscrim-
inantisnon-negativeforasolutiontoexist,orthatthediscriminantisaperfect
square(inorderforthesolutionstobeaninteger).Thisapproach,however,has
limitedusandmaysometimesleadtoadeadend.
Anotherapproachwouldbetotrytofactorithegivenexpressiontotrytoderive
ufuldivisibilitypropertiesorinequalitiesthatwillallowyoutoreducethescope
ofconsiderationtojustafewcas.Factorisationisalwaysoneapproachthatyou
shouldkeepinmind,asitmaysometimesyieldsurprisingresults.
Alternatively,toshowthatacertainequationhasnointegersolutions,wecanalso
uFermat’smethodofinfinitedescent.Toshowthatanequationinvolvingper-
fectsquareshasinfinitelymanysolutions,wemaywanttoshowthattheequation
canbereducedtofindingthesolutionstopythagoreantriples,forwhichtherewill
beinfinitelymanysolutions.Thetechniqueswillnotbecoveredindetailinthis
lesson.
2.Provethatacertainexpressionisalwaysaperfectsquare(orperfectcube).
Themoststraightforwardwaytodothisistoshowthatitcanalwaysbefactorid
intoaperfectsquare(orcube)!Thefollowingresultmaybeuful:Ifabis
aperfectsquareandgcd(a,b)=1,thenaandbmustalsobeperfectsquares.
Anotherstandardapproachistoprovebycontradiction.Suppotheexpression
isnotaperfectsquare,andshowthatitleadstoacontradiction.
Ifthequestionisofthetype“provethat(someexpression)isalwaysaperfect
squareforallvaluesofn”,thenyoumaywanttoconsiderusingmathematical
induction.
3.Provethatacertainexpressionisneveraperfectsquare.Simpleproblems
ofthissortcanbedonebyconsideringtheresidueoftheexpressionmoduloa
suitablenumber,sayn.Ifitisnotaquadraticresidueofn,thenitcannotbea
perfectsquare.
Anothercommontechniqueinvolvesboundingbetweenconcutivesquares(or
concutivecubes).Ifn<x<(n+1)foranintegern,thenxcannotbeaperfect
22
square.
Alternatively,ifweareabletoshowthataprimeporanoddpoweroftheprime,
p,dividestheexpressionexactly,thentheexpressioncannotbeaperfect
2k+1
square.
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4.Aprobleminvolvingtwogivenexpressionsthatarebothperfectsquares.If
youaregivenexpressionAandexpressionBandyouaretoldthattheyareboth
perfectsquares,thenyoumightwishtoask,whatconditionsmustholdforboth
ofthemtobeperfectsquares?IfweletA=mandB=n,itmaybeufulto
22
considerA−B=(m+n)(m−n).Itmayalsohelptoconsiderifmornmustbe
oddoreven(bytakingAandBmodulo4,forinstance).Consideringresidues
oftheexpressionsmoduloasuitablenumbermaysometimesalsoleadyoutoa
solution(oracontradiction).
Iftheproblemwantsyouto“findallintegersxsuchthatAandBarebothperfect
squares”,whereAandBareexpressionsinvolvingx,youmightwanttofinda
boundforx.Forinstance,againwesuppoA=mandB=n.Thenyoumay
22
wanttoshowthatifxexceedsacertainnumber,thenm<n<(m+1),thusA
222
andBcannotbothbeperfectsquares,acontradiction.
AnotherapproachtothiskindofproblemistotrytoobtainaPell’sequationfrom
thegivenexpressions.However,thatisbeyondthescopeofourlesson.
5.EquationswithunknownexponentsIftheequationhastermswithanunknown
exponente.g.7,thenwemaywishtoconsiderresiduesmoduloasuitablenumber
x
to‘eliminate’theterm.Forinstance,wecanumod7togetridof7.Alterna-
x
tively,mod3andmod8arealsoufulsince7≡1(mod3)and7≡±1(mod
xx
8).Thismightleadustosomethinguful.Alternatively,iftherearetwoormore
termswithunknownexponents,suchas3+4=5,thenconsideringmod3and
xyz
mod4respectively,youwillobtainxandzareeven.Theequationcanthenbe
factoridasadifferenceofsquares.
ResiduesofSquares,CubesandFourthPowers
Thefollowingresultsforperfectsquaresareeasytoverify:
•x≡0,1(mod4).
2
Moreprecily,x≡0(mod4)⇔xiseven,andx≡1(mod4)⇔xisodd.
22
•x≡0,1,4(mod8).
2
Inparticular,notethatx≡1(mod8)⇔xisodd.
2
•x≡0,1,4,9(mod16)
2
•x≡0,1(mod3)
2
•x≡0,1,4(mod5)
2
•x≡0,1,2,4(mod7)
2
Herearesomeufulpropertiesofperfectcubes:
•x≡0,±1(mod7)
3
•x≡0,±1(mod9)
3
Finally,forperfectfourthpowers:
•x≡0,1(mod5)
4
•x≡0,1(mod16)
4
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Example1.Findallpairsofprimenumbers(p,q)suchthat
(RussiaMO)
p−q=(p+q).
352
Example2.Letdbeanypositiveintegerthatisnot2,5or13.Provethatatleastoneof
thenumbers2d−1,5d−1,13d−1isnotaperfectsquare.
Example3.Findallnon-negativeintegersolutionstotheequation3−y=1.
x3
Example4.Provethattherearenointegersxandysuchthat
(Putnam1954)
x+3xy−2y=122.
22
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ClassroomProblems
1.Letnbeanaturalnumbersuchthat2n+1and3n+1arebothperfectsquares.
Provethat5n+3iscomposite.
2.If2n+1and3n+1arebothperfectsquares,provethatnmustbedivisibleby40.
3.Determineallprimespsuchthat5+12isaperfectsquare.
pp
√
4.Letnbeaninteger.Provethatif228n+1+2isaninteger,thenitisaperfect
2
square.
5.Findallnaturalnumbersnsuchthat2+2+2isaperfectsquare.
811n
6.Provethatthesystemofequations
x+6y=z
222
6x+y=t
222
hasnonon-trivialsolutions.
7.Findallintegersxandysuchthatx+3yandy+3xarebothperfectsquares.
22
8.Findallintegersxandysuchthatx+y,x+2yand2x+yareallperfectsquares.
4442009
+x+...+x=2009.9.Determineallintegersolutionstotheequationx
128
10.Findallnon-negativeintegers(x,y)satisfying(xy−7)=x+y.
222
11.Provethattheequationy=x−4hasnointegersolutions.
25
12.Findallpairsofintegers(x,y)satisfyingtheequation
(SMO(S)2004/Round2)
(x+y)=1+16y.
222
13.Showthattheequation2−1=zhasnointegersolutionsifx,m>1.
xm
14.Dothereexistintegersxandysuchthat19=x+y?Justify
(SMO(O)1998/B3)
1934
youranswer.
15.Findallpairsofnonnegativeintegers(x,y)satisfying
(NTST2007)
(14y)+y=2007.
xx+y
16.Whatisthesmallestpositiveintegernsuchthatthereexists
(IMOshortlist2002)
3332002
+x+...+x=2002?integersx,x,...,xsatisfyingx
12n
12
n
17.Determineallpairs(x,y)ofintegerssuchthat
(IMO2006)
1+2+2=y.
x2x+12
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