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Question Number 134981 by oooooooo last updated on 09/Mar/21

(a+1)(b+1)(c+1)=2abc a,b,c ε N

(a+1)(b+1)(c+1)=2abca,b,cεN

Answered by MJS_new last updated on 10/Mar/21

let a≤b≤c (1) c=(a+1)(b+1) ⇒ b=((a+2)/(a−1))∧c=(((a+1)(2a+1))/(a−1)) a=2∧b=4∧c=15 (2) ab=c+1 ⇒ b=((a+3)/(a−1))∧c=(((a+1)^2 )/(a−1)) a=2∧b=5∧c=9 a=b=3∧c=8 (3) bc=(a+1)(b+1) ⇒ b=((a+1)/(a−2))∧c=2a−1 a=3∧b=4∧c=5 (4) b=2a+2∧c=2a+3 ⇔ c=b+1∧2b=c+1 [the idea here is to get an additional factor 2 on the lhs] a=2∧b=6∧c=7

leta⩽b⩽c(1)c=(a+1)(b+1)⇒b=a+2a−1∧c=(a+1)(2a+1)a−1a=2∧b=4∧c=15(2)ab=c+1⇒b=a+3a−1∧c=(a+1)2a−1a=2∧b=5∧c=9a=b=3∧c=8(3)bc=(a+1)(b+1)⇒b=a+1a−2∧c=2a−1a=3∧b=4∧c=5(4)b=2a+2∧c=2a+3⇔c=b+1∧2b=c+1[theideahereistogetanadditionalfactor2onthelhs]a=2∧b=6∧c=7

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