Download Consecutive Integer Squares

Consecutive Integer Squares
11October2016
StephenMorris,Newbury,Berkshire,UK
[email protected]
FromStanWagon’sProblemoftheWeek1229
ProblemStatement:
For which positive integers n is there a sum of n consecutive integers that is a perfect square?
Solution:
Anaturalapproachistocalculatesuchasumasanexplicitformula.Howeverweprefertoreframe
theproblemasoneaboutrectanglesandintegertripleswhichavoidstheneedformuchalgebra.
Wewilluseonenoteofgeometryandsomesimplelogicasourmaintools.
ThecompletegeneralsolutionisgivenbelowasTheorem1.Theproofwillcomesimplyfromour
approachtowardstheend.
Theorem1:Anygivenpositiveintegernfactorisesuniquelyas2fg2hwhereg,hareoddandhis
squarefree.Thegeneralsolutionforsequencesofconsecutiveintegersoflengthnthatsumtoa
perfectsquareisgivenby:
hr2–(n-1)/2..hr2+(n-1)/2
(hr2–(n-1))/2..(hr2+(n-1))/2
Non-existent
foranyintegerr,
whenf=0
foranyoddintegerr, whenfodd
whenf>0even
Considerarectanglewithwidthintegera,andheightintegerb>a.Therectangleissuchthatitcan
besplitintoidenticalhalvesusingasteppedlinecomposedofedgesoftheunitsquares.
Thecentreoftherectangleisacentreofrotationforthesteppedline,andsoisinthemiddleofan
edgeofaunitsquare.Ifthisisahorizontaledge(asshown)thenaisoddandbeven.Ifitisvertical
thenaisevenandbodd.Socertainlyaandbhavedifferentparity.
Eachhalfrepresentsasequenceofconsecutivenumbers.Iftheareais2c2thenitisoursolutionto
theoriginalproblemforn=a.
Lessobviousisthatitisalsoasolutionforn=b,ifweallowthesequencetocontainzeroand
negativeintegers.Weillustratethiswithanexample:
Inthesecondcase,ontherightofthediagram,theareaabovetherectangleisequaltothearea
below.Sincetheformercontributesapositiveamounttothesumandthelattercontributesa
negativeamounttheycancelout.Sothesumofthesequenceishalfoftheareaoftherectangle.
Anysequenceofconsecutiveintegerswillgeneratesucharectangle.
Withthisobservationwemayreframetheproblemasfollows.
Wesaythatatripleofpositiveintegers,(a,b,c),isasolutionifitsatisfiesthefollowingconditions:
1. aisodd
2. biseven
3. ab=2c2
Notethatwehaveremovedtheconditionthatb>a.
Weusethetermsolutionsinceitrepresentssolutionstotheoriginalproblemforn=aandn=b.
Theexampleaboveisthesolution(3,6,3)andsorepresentsasolutiontotheoriginalproblemfor
n=3andn=6.
OurnextresultsareputintheformofaLemma.
Lemma1:
i.
ii.
iii.
iv.
v.
(a,2a,a)isasolutionforanyoddpositiveintegera.
If(a,b,c)isasolutionthen
a. (a,r2b,rc)isasolutionforanyintegerr,
b. (r2a,b,rc)isasolutionforanyoddintegerr.
Inanysolutionbisdivisibleby2anoddnumberoftimes.
Thereisasolutionfornpreciselywheneithernisoddornisdivisibleby2anoddnumberof
times.
Ifthereisasolutionfornthenthereisasolutionwheretheconsecutiveintegersarepositive.
ProofofLemma1:
i.andii.aresimpletocheckagainsttheconditionsforasolution.
iii.followsfromcondition3,thatab=2c2.WeconsiderthattheRHSisdivisibleby2anoddnumber
oftimesandthataisodd.
Foriv:Ifnisoddweusei.Ifniseventhenweusei.andii.a.withrapowerof2.Thesituation
wherenisdivisibleby2anevennumberoftimesisexcludedbyiii.
Forv.weuseii.togetarectanglewithntheshortestside.
Lemma1.ivsolvesthestatedproblem.Lemma1.vshowsthatweonlyneedconsiderpositive
integers.
Theabovedemonstrateswhenasolutionexistsforn.Wewishtocalculatethesequences
generatedby(a,b,c).Therearetwosuchsequences,forn=aandn=b.
Letsbethefirstnumberinasequenceandtbethelastnumberinthesequencesothatthe
sequenceiss..t.
Firsttakethatcasewheren=a.Wehavethefollowingequationstosolve.
t–s=a–1
sincethelengthofthesequenceisn=a
t+s=b
bytheconstructionoftherectangle
fromwhich
t=(b+a–1)/2
s=(b–a+1)/2
Intheothercasewheren=bthenwemustswitchthevaluesaandbtoget
t=(a+b–1)/2
s=(a–b+1)/2
Notethattisthesameinbothcases.Thedifferenceinsisexplainedbyobservingthatweadd
equivalentnegativeandpositiveintegers,pluszero,tothestartofthesequence.
Applyingthistoourexample(3,6,3)wegett=(6+3-1)/2=4.Forn=3wegets=(6-3+1)/2=2.For
n=6wegets=(3-6+1)/2=-1.Sothesequencesare:
2+3+4=9
-1+0+1+2+3+4=9
AWorkedExample
Toshowthatthisisaconstructivesolutionletusconsideraworkedexample.
Question:Canwefindasequenceof40consecutivepositiveintegerssuchthatthesumisaperfect
square?
Answer:
40is235,sowestartbyusingLemma1.itoget(5,10,5).
NextweuseLemma1.ii.atoget(5,40,10).
Aswewantapositivesolutionwemusthave40thesmallestofaandb,soweuseLemma1.ii.bto
get(45,40,30).
Usingourformulaeforsandtwefindthesequenceis(45-40+1)/2..(45+40-1)/2whichis3..42.
Weshouldcheckthatthesumis302.
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𝑖=
!!!
=
!
𝑖
𝑖 −
!!!
!!!
42 42 + 1
2 2+1
−
2
2
= 903 − 3
= 30! TheGeneralSolution
Anypositiveintegernwithasolutioncanbefactoriseduniquelyasn=2fg2hwherefiszeroorodd,g
andhareoddandhissquare-free.
Firstweconsiderthecasewherenisodd,whichoccurswhenf=0andson=g2h.
Firstwefindthegeneralsolutionintermsoftriples.Atriplerepresentingasolutionmusthavea=n
=g2h,sincenisodd.Thenbmustbedivisibleby2htosatisfyab=2c2.Thatgivesthegeneral
solution(g2h,2r2h,rgh)foranyintegerr.
Thesequenceisgivenby
s=(b–a+1)/2=(2r2h-g2h+1)/2=hr2–(n-1)/2
t=(b+a-1)/2=(2r2h+g2h-1)/2=hr2+(n-1)/2
Sothesequenceis
hr2–(n-1)/2..hr2+(n-1)/2
Intheothercaseniseven,whichoccurswhenf>0.FromtheLemma1weknowthatfmustbeodd.
Againwefindthegeneralsolutionintermsoftriples.Atriplerepresentingasolutionmusthaveb=
n=2fg2h,sinceniseven.Thenamustbedivisiblebyhtosatisfyab=2c2.Wehavethegeneral
solution(r2h,2fg2h,2(f-1)/2rgh)foranyoddintegerr.
Thesequenceisgivenby
s=(a-b+1)/2=(r2h–2fg2h+1)/2=(hr2–(n-1))/2
t=(a+b-1)/2=(r2h+2fg2h-1)/2=(hr2+(n-1))/2
Sothesequenceis
(hr2–(n-1))/2..(hr2+(n-1))/2
ThisprovesTheorem1whichwerestatehereforconvenience.
Theorem1:Anygivenpositiveintegernfactorisesuniquelyas2fg2hwhereg,hareoddandhis
squarefree.Thegeneralsolutionforsequencesofconsecutiveintegersoflengthnthatsumtoa
perfectsquareisgivenby:
hr2–(n-1)/2..hr2+(n-1)/2
(hr2–(n-1))/2..(hr2+(n-1))/2
Non-existent
foranyintegerr,
whenf=0
foranyoddintegerr, whenfodd
whenf>0even
ATableofSolutionsforsmalln
UsingtheformulaeinTheorem1wecanwriteatableofpossiblesolutionsforeachn.
n
1
2
3
4
f
0
1
0
2
g
1
1
1
1
h
1
1
3
1
From
r2
(r2-1)/2
3r2-1
To
r 2
(r2+1)/2
3r2+1
SqrtSum
r
r
3r
5
6
7
8
9
10
11
12
0
1
0
3
0
1
0
2
1
1
1
1
3
1
1
1
5
3
7
1
1
5
11
3
5r2-2
(3r2-5)/2
7r2-3
(r2-7)/2
r2-4
(5r2-7)/2
11r2-5
5r2+2
(3r2+5)/2
7r2+3
(r2+7)/2
r2+4
(5r2+7)/2
11r2+5
5r
3r
7r
2r
3r
5r
11r
13
14
15
16
0
1
0
4
1
1
1
1
13
7
15
1
13r2-6
(7r2-13)/2
15r2-7
13r2+6
(7r2+13)/2
15r2+7
13r
7r
15r
17
18
0
1
1
3
17
1
17r2-8
(r2-17)/2
17r2+8
(r2+17)/2
17r
3r
ris…
Odd
None
Exist
Odd
Odd
Odd
None
Exist
Odd
None
Exist
Odd
19
20
0
2
1
1
19
5
19r2-9
19r2+9
21
22
0
1
1
1
21
11
21r2+10
21r
(11r2+21)/2 11r
23
24
25
26
0
3
0
1
1
1
5
1
23
3
1
13
23r2+11
(3r2+23)/2
r2+12
(13r2+25)/2
23r
6r
5r
13r
Odd
Odd
27
28
0
2
3
1
3
7
21r2-10
(11r221)/2
23r2-11
(3r2-23)/2
r2-12
(13r225)/2
3r2-13
None
Exist
Odd
3r2+13
9r
29
30
0
1
1
1
29
15
29r2+14
29r
2
(15r +29)/2 15r
31
32
33
34
0
5
0
1
1
1
1
1
31
1
33
17
29r2-14
(15r229)/2
31r2-15
(r2-31)/2
33r2-16
(17r233)/2
None
Exist
Odd
31r2+15
(r2+31)/2
33r2+16
(17r2+33)/2
Odd
Odd
19r
31r
4r
33r
17r