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Question Number 189624 by Rupesh123 last updated on 19/Mar/23

Answered by Rasheed.Sindhi last updated on 19/Mar/23

x^2 −(a+b)x+ab=x^2 +(c−6)x−6c+5 −(a+b)=c−6 ∧ ab=5−6c c=−a−b+6 ∧ ab=5−6(−a−b+6) ab=6a+6b−31 ab−6a=6b−31 a=((6b−31)/(b−6))=((6(b−6)+5)/(b−6))=6+(5/(b−6))∈Z b−6=±1,±5 b=±1+6,±5+6 ⇒b=1,5,7,11 a=5,1,11,7 c=0,0,−12,−12 (a,b,c)=(1,5,0),(5,1,0),(7,11,−12),(11,7,−12) 4 triples

x2−(a+b)x+ab=x2+(c−6)x−6c+5−(a+b)=c−6∧ab=5−6cc=−a−b+6∧ab=5−6(−a−b+6)ab=6a+6b−31ab−6a=6b−31a=6b−31b−6=6(b−6)+5b−6=6+5b−6∈Zb−6=±1,±5b=±1+6,±5+6⇒b=1,5,7,11a=5,1,11,7c=0,0,−12,−12(a,b,c)=(1,5,0),(5,1,0),(7,11,−12),(11,7,−12)4triples

Commented by Rupesh123 last updated on 19/Mar/23

Excellent!

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