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Question Number 191028 by Mingma last updated on 16/Apr/23
Answered by Frix last updated on 16/Apr/23
Notaproofbutanideaforaproof:(n−1)n(n+1)×N=k3;k∈NWeknowthat1.gcd(n−1,n)=12.gcd(n,n+1)=13.gcd(n−1,n+1)∈{1,2}⇒(n−1)n(n+1)isnotaperfectcube⇒Nmusthaveenoughprimefactorstocompletetheperfectcube.ObviouslythisisimpossibleforN=n±2LetN∈Nthenecessaryfactortogetk3n=2N=36n=3N=9n=4N=450n=5N=225n=6N=44100n=7N=1764n=8N=147...
Answered by a.lgnaoui last updated on 16/Apr/23
soienta(a+1)(a+2)et(a+3)les4nombresentierssuccessivesalirsa4+6a3+11a2+6a=a3(a+6+11a+6a3)onremarqueque11a+6a3=1a(11+6a2)E=a3[a+6+1a(11+6a2)]maxvaldurpossible(a=1)(7+17)a3nest[pascubeparfait(2a)3<24a3<(3a)3donc∄a/a∈Npourlequela(a+1)(a+2)(a+3)estcubeparfait∙2−suposonsquea+1=λ3alorsa=λ3−1=(λ−1)(λ2+λ+1)nestpauncubeparfaitmemecaspoura+2eta+3respectivementparconsequent⇒∃!untermeai/ai3noncubeparfait1<i<4∏i=1i=4ai(ai+1=ai+1entier):nestpascubeparfaitalorslepriduitde4entierssuccessivesnepeutpasetreuncubeparfait
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