Question Number 162787 by amin96 last updated on 01/Jan/22
$$\boldsymbol{\mathrm{happy}}\:\boldsymbol{\mathrm{new}}\:\boldsymbol{\mathrm{year}} \\ $$$$\left\{\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\right\}\in\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}}\:\:\:\: \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{b}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\mathrm{1}\right)=? \\ $$
Answered by alephzero last updated on 01/Jan/22
$${p}\left({a}\right)\:=\:{aa}^{\mathrm{2}} \:+\:{ba}\:+\:{c}\:=\:{a}^{\mathrm{3}} \:+\:{ba}\:+\:{c}\:=\:\mathrm{0} \\ $$$${p}\left({b}\right)\:=\:{ab}^{\mathrm{2}} \:+\:{bb}\:+\:{c}\:=\:{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}\:=\:\mathrm{0} \\ $$$${p}\left(\mathrm{1}\right)\:=\:{a}^{\mathrm{2}} \:+\:{b}\:+\:{c}\:=\:? \\ $$$$\begin{cases}{{a}^{\mathrm{3}} +\:{ba}\:+\:{c}\:=\:\mathrm{0}}\\{{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}\:=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{a}^{\mathrm{3}} \:+\:{ba}\:=\:−{c}}\\{{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:−{c}}\end{cases}\: \\ $$$$\Rightarrow\:{a}^{\mathrm{3}} \:+\:{ba}\:=\:{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}\left({a}^{\mathrm{2}} \:+\:{b}\right)\:=\:{b}^{\mathrm{2}} \left({a}\:+\:\mathrm{1}\right) \\ $$$$\left({a}_{\mathrm{1}} ;\:{b}_{\mathrm{1}} \right)\:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}};\:−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left({a}_{\mathrm{2}} ;\:{b}_{\mathrm{2}} \right)\:=\:\left(−\mathrm{2};\:\mathrm{4}\right) \\ $$$$\left({a}_{\mathrm{3}} ;\:{b}_{\mathrm{3}} \right)\:=\:\left(\mathrm{0};\:\mathrm{0}\right) \\ $$$$\mathrm{But}\:\left\{{a};\:{b};\:{c}\right\}\:\in\:\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\Rightarrow\:\left({a};\:{b}\right)\:=\:\left(−\mathrm{2};\:\mathrm{4}\right) \\ $$$$\Rightarrow\begin{cases}{−\mathrm{2}^{\mathrm{3}} \:+\:\left(−\mathrm{2}\right)\:×\:\mathrm{4}\:=\:−{c}}\\{−\mathrm{2}\:×\:\mathrm{4}^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \:=\:−{c}}\end{cases} \\ $$$$\begin{cases}{−\mathrm{8}\:+\:\left(−\mathrm{8}\right)\:=\:−{c}}\\{−\mathrm{36}\:+\:\mathrm{16}\:=\:−{c}}\end{cases} \\ $$$$\Rightarrow\:−\mathrm{16}\:=\:−{c} \\ $$$$\Rightarrow\:{c}\:=\:\mathrm{16} \\ $$$${a}^{\mathrm{2}} \:+\:{b}\:+\:{c}\:=\:\left(−\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}\:+\:\mathrm{16}\:=\:\mathrm{4}\:+\:\mathrm{4}\:+ \\ $$$$+\:\mathrm{16}\:=\:\mathrm{8}\:+\:\mathrm{16}\:=\:\mathrm{24} \\ $$$${p}\left(\mathrm{1}\right)\:=\:\mathrm{24} \\ $$
Commented by mr W last updated on 01/Jan/22
$${solution}\:{is}\:{not}\:{unique}. \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +{x}−\mathrm{2}\:{is}\:{also}\:{ok}. \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0}. \\ $$$${in}\:{fact}\:{any}\:{a}={b}={n},\:{c}=−{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\:{with} \\ $$$${n}\in{Z}\:{is}\:{ok}. \\ $$