Épreuve de setion européenne
Perfet numbers
Aperfetnumberisanaturalnumberthatisthesumofitsproperdivisors(alldivisorsexept
thenumberitself). Forexample,28is perfet. InTheElements,bookIX,proposition36,Eulid
wrote:
If asmanynumbersas weplease, beginningfrom a unit,are set outontinuously in
double proportion,untilthesumofallbeomesprime,andifthesummultipliedinto
thelastmakessomenumber,theprodutwillbeperfet.
With theadvantages of modernnotation, we anexpresswhat Eulid meantmorepreisely:
ifwebeginwith 1andadd toitsuessivelyhigherpowersof2sotheresultingsum
1 + 2 + 4 + 8 + . . . + 2 n is aprime number,thenthe numberN n = 2 n (1 + 2 + 4 + 8 + . . . + 2 n )
, formed by
multiplyingthesum
1 + 2 + 4 + 8 + . . . + 2 n byitslast summand 2 n,mustbeperfet.
Notein passingthat the numbers
K n of the form 1 + 2 + 4 + 8 + . . . + 2 n don'tneed to be
primeat all. Eulid'sperfetnumbertheorem appliedonlytothosespeialaseswherethissum
indeedturnsouttobeaprime.
WeshallnotlookatEulid'sproofofthisresultbutshallinsteadonsideraspeiexample.
Forinstane,
1 + 2 + 4 + 8 + 16 = 31
,aprime. Then,N = 16 × 31 = 496
should beperfet. Toseethat itis, welistalltheproperdivisorsof496namely, 1,2,4,8,16, 31,62, 124and248
andaddthemtoget496,aspromised.
AdaptedfromJourney through genius, thegreattheoremsof mathematis,byWilliamDunham
Questions
1. Usethetexttoexplainthefollowingwords: primenumber,properdivisor,perfetnumber.
2. Chekthat28is perfet.
3. Find asimplerformulafor
K n anddeduethegeneralexpressionofN n.
4. In orderto hekthat numbers
K n of theform 1 + 2 + 4 + 8 + . . . + 2 n don't needto be
prime atall',ompleteatable ofvaluesof
K n,forn
rangingfrom1to5.
5. Chekthat
N 3isnotperfet.
6. ProofofEulid'stheorem: Wesupposethat
n
isanaturalnumber.(a) Listallthedivisorsof
2 n.
(b) Henededuealltheproperdivisorsof
N n = 2 n × K n,whereK n = 2 n+1 − 1
isprime.
() Add up all the properdivisors found in the previous question and provethat