Question Number 204350 by universe last updated on 14/Feb/24
Answered by mr W last updated on 14/Feb/24
$${P}\left({a}\right)={a}^{\mathrm{4}} +{a}^{\mathrm{2}} {b}+{ac}+{d}=\mathrm{1}\:\:\:\:…\left({i}\right) \\ $$$${P}\left({b}\right)={ab}^{\mathrm{3}} +{b}^{\mathrm{3}} +{bc}+{d}=−\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${a}\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)+{b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}\left({a}−{b}\right)=\mathrm{2} \\ $$$${a}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)+{b}\left({a}+{b}\right)+{c}=\frac{\mathrm{2}}{{a}−{b}}={whole}\:{number} \\ $$$$\Rightarrow{a}−{b}=\pm\mathrm{1},\:\pm\mathrm{2} \\ $$$$\Rightarrow\mid{a}−{b}\mid_{{max}} =\mathrm{2}\:\checkmark \\ $$
Commented by universe last updated on 14/Feb/24
$$\:{but}\:{sir}\:{how}\:{it}\:{is}\:{possible}\:{p}\left({b}\right)=−\mathrm{1} \\ $$$${because}\:{a},{b},{c},{d}\:{all}\:{are}\:{positive}\:{integer}\:{so} \\ $$$${ab}^{\mathrm{3}} +{b}^{\mathrm{3}} +{bc}+{d}\:\geqslant\mathrm{0}\:{i}\:{think}\:{question}\:{is}\:{wrong}?? \\ $$
Commented by mr W last updated on 14/Feb/24
$${i}\:{think}\:{the}\:{question}\:{means}\:{just}\: \\ $$$${integer}\:{numbers}\:\left(\mathbb{Z}\right). \\ $$$${in}\:{different}\:{countries}\:{it}\:{may}\:{have} \\ $$$${different}\:{meanings}. \\ $$
Commented by mr W last updated on 14/Feb/24
Commented by universe last updated on 14/Feb/24
$${thank}\:{you}\:{sir} \\ $$