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Question-189624

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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
$${x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}={x}^{\mathrm{2}} +\left({c}−\mathrm{6}\right){x}−\mathrm{6}{c}+\mathrm{5} \\ $$$$−\left({a}+{b}\right)={c}−\mathrm{6}\:\:\wedge\:{ab}=\mathrm{5}−\mathrm{6}{c} \\ $$$${c}=−{a}−{b}+\mathrm{6}\:\wedge\:{ab}=\mathrm{5}−\mathrm{6}\left(−{a}−{b}+\mathrm{6}\right) \\ $$$${ab}=\mathrm{6}{a}+\mathrm{6}{b}−\mathrm{31} \\ $$$${ab}−\mathrm{6}{a}=\mathrm{6}{b}−\mathrm{31} \\ $$$${a}=\frac{\mathrm{6}{b}−\mathrm{31}}{{b}−\mathrm{6}}=\frac{\mathrm{6}\left({b}−\mathrm{6}\right)+\mathrm{5}}{{b}−\mathrm{6}}=\mathrm{6}+\frac{\mathrm{5}}{{b}−\mathrm{6}}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:{b}−\mathrm{6}=\pm\mathrm{1},\pm\mathrm{5} \\ $$$$\:\:\:\:\:{b}=\pm\mathrm{1}+\mathrm{6},\pm\mathrm{5}+\mathrm{6} \\ $$$$\Rightarrow{b}=\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11}\:\: \\ $$$$\:\:\:{a}=\mathrm{5},\mathrm{1},\mathrm{11},\mathrm{7} \\ $$$$\:\:\:{c}=\mathrm{0},\mathrm{0},−\mathrm{12},−\mathrm{12} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{1},\mathrm{5},\mathrm{0}\right),\left(\mathrm{5},\mathrm{1},\mathrm{0}\right),\left(\mathrm{7},\mathrm{11},−\mathrm{12}\right),\left(\mathrm{11},\mathrm{7},−\mathrm{12}\right) \\ $$$$\mathrm{4}\:{triples} \\ $$
Commented by Rupesh123 last updated on 19/Mar/23
Excellent!

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