Question Number 189624 by Rupesh123 last updated on 19/Mar/23
Answered by Rasheed.Sindhi last updated on 19/Mar/23
$${x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}={x}^{\mathrm{2}} +\left({c}−\mathrm{6}\right){x}−\mathrm{6}{c}+\mathrm{5} \\ $$$$−\left({a}+{b}\right)={c}−\mathrm{6}\:\:\wedge\:{ab}=\mathrm{5}−\mathrm{6}{c} \\ $$$${c}=−{a}−{b}+\mathrm{6}\:\wedge\:{ab}=\mathrm{5}−\mathrm{6}\left(−{a}−{b}+\mathrm{6}\right) \\ $$$${ab}=\mathrm{6}{a}+\mathrm{6}{b}−\mathrm{31} \\ $$$${ab}−\mathrm{6}{a}=\mathrm{6}{b}−\mathrm{31} \\ $$$${a}=\frac{\mathrm{6}{b}−\mathrm{31}}{{b}−\mathrm{6}}=\frac{\mathrm{6}\left({b}−\mathrm{6}\right)+\mathrm{5}}{{b}−\mathrm{6}}=\mathrm{6}+\frac{\mathrm{5}}{{b}−\mathrm{6}}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:{b}−\mathrm{6}=\pm\mathrm{1},\pm\mathrm{5} \\ $$$$\:\:\:\:\:{b}=\pm\mathrm{1}+\mathrm{6},\pm\mathrm{5}+\mathrm{6} \\ $$$$\Rightarrow{b}=\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11}\:\: \\ $$$$\:\:\:{a}=\mathrm{5},\mathrm{1},\mathrm{11},\mathrm{7} \\ $$$$\:\:\:{c}=\mathrm{0},\mathrm{0},−\mathrm{12},−\mathrm{12} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{1},\mathrm{5},\mathrm{0}\right),\left(\mathrm{5},\mathrm{1},\mathrm{0}\right),\left(\mathrm{7},\mathrm{11},−\mathrm{12}\right),\left(\mathrm{11},\mathrm{7},−\mathrm{12}\right) \\ $$$$\mathrm{4}\:{triples} \\ $$
Commented by Rupesh123 last updated on 19/Mar/23
Excellent!