Question Number 830 by prakash jain last updated on 20/Mar/15
$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={f}\left({x}\right){f}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{Find}\:{f}\left({x}\right)\:\mathrm{such}\:\mathrm{that}\:{f}\left({x}\right)\neq\mathrm{0}. \\ $$
Commented by 123456 last updated on 23/Mar/15
$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$${f}\left({x}^{\mathrm{2}} \right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{\mathrm{2}{i}} \equiv\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{b}_{{i}} {x}^{{i}} \\ $$$${f}\left({x}−\mathrm{1}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \left({x}−\mathrm{1}\right)^{{i}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{c}_{{i}} {x}^{{i}} \\ $$$${f}\left({x}\right){f}\left({x}−\mathrm{1}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} {x}^{{i}} \underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{c}_{{i}} {x}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{d}_{{i}} {x}^{{i}} \\ $$$${b}_{{i}} =\begin{cases}{{a}_{{i}/\mathrm{2}} }&{{i}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)}\\{\mathrm{0}}&{{i}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)}\end{cases} \\ $$$$\left({x}−\mathrm{1}\right)^{{i}} =\underset{{j}=\mathrm{0}} {\overset{{i}} {\sum}}\begin{pmatrix}{{i}}\\{{j}}\end{pmatrix}\:{x}^{{j}} \left(−\mathrm{1}\right)^{{i}−{j}} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \left({x}−\mathrm{1}\right)^{{i}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \underset{{j}=\mathrm{0}} {\overset{{i}} {\sum}}\begin{pmatrix}{{i}}\\{{j}}\end{pmatrix}\:{x}^{{j}} \left(−\mathrm{1}\right)^{{i}−{j}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{c}_{{i}} {x}^{{i}} \\ $$$${c}_{{i}} \overset{?} {=}\underset{{j}={i}} {\overset{{n}} {\sum}}\begin{pmatrix}{{j}}\\{{i}}\end{pmatrix}\left(−\mathrm{1}\right)^{{j}−{i}} {a}_{{j}} \\ $$$${d}_{{i}} =\underset{{j}=\mathrm{0}} {\overset{{i}} {\sum}}{a}_{{j}} {c}_{{i}−{j}} \\ $$$${d}_{{i}} ={b}_{{i}} \\ $$
Commented by 123456 last updated on 23/Mar/15
$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}+\boldsymbol{{b}} \\ $$$${f}\left({x}−\mathrm{1}\right)={a}\left({x}−\mathrm{1}\right)+{b}={ax}+\left({b}−{a}\right) \\ $$$${f}\left({x}^{\mathrm{2}} \right)={ax}^{\mathrm{2}} +{b} \\ $$$${ax}^{\mathrm{2}} +{b}={a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}\left(\mathrm{2}{b}−{a}\right){x}+{b}\left({b}−{a}\right) \\ $$$$\begin{cases}{{a}={a}^{\mathrm{2}} }&{{a}=\mathrm{0}\vee{a}=\mathrm{1}}\\{\mathrm{0}={a}\left(\mathrm{2}{b}−{a}\right)}&{{a}=\mathrm{0}\vee\mathrm{2}{b}={a}}\\{{b}=\left({b}−{a}\right){b}}&{{b}=\mathrm{0}\vee{b}−{a}=\mathrm{1}}\end{cases} \\ $$$${a}=\mathrm{0}\Rightarrow\begin{cases}{\mathrm{0}=\mathrm{0}}&{}\\{\mathrm{0}=\mathrm{0}}&{}\\{\boldsymbol{{b}}=\boldsymbol{{b}}^{\mathrm{2}} }&{\boldsymbol{{b}}=\mathrm{0}\vee\boldsymbol{{b}}=\mathrm{1}}\end{cases} \\ $$$${a}=\mathrm{1}\Rightarrow\begin{cases}{\mathrm{1}=\mathrm{1}}&{}\\{\mathrm{0}=\mathrm{2}\boldsymbol{{b}}−\mathrm{1}}&{\boldsymbol{{b}}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\boldsymbol{{b}}=\left(\boldsymbol{{b}}−\mathrm{1}\right)\boldsymbol{{b}}}&{\boldsymbol{{b}}=\mathrm{0}\vee\boldsymbol{{b}}=\mathrm{2}}\end{cases} \\ $$$${a}=\mathrm{2}{b}\Rightarrow\begin{cases}{\mathrm{2}\boldsymbol{{b}}=\mathrm{4}\boldsymbol{{b}}^{\mathrm{2}} }&{\boldsymbol{{b}}=\mathrm{0}\vee\boldsymbol{{b}}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{0}=\mathrm{0}}&{}\\{\boldsymbol{{b}}=−\boldsymbol{{b}}^{\mathrm{2}} }&{\boldsymbol{{b}}=\mathrm{0}\vee\boldsymbol{{b}}=−\mathrm{1}}\end{cases} \\ $$$${a}={b}−\mathrm{1}\Rightarrow\begin{cases}{\boldsymbol{{b}}−\mathrm{1}=\left(\boldsymbol{{b}}−\mathrm{1}\right)^{\mathrm{2}} }&{\boldsymbol{{b}}=\mathrm{1}\vee\boldsymbol{{b}}=\mathrm{2}}\\{\mathrm{0}=\left(\boldsymbol{{b}}−\mathrm{1}\right)\left(\boldsymbol{{b}}+\mathrm{1}\right)}&{\boldsymbol{{b}}=−\mathrm{1}\vee\boldsymbol{{b}}=\mathrm{1}}\\{\boldsymbol{{b}}=\boldsymbol{{b}}}&{}\end{cases} \\ $$$${f}\left({x}\right)=\mathrm{0}\vee{f}\left({x}\right)=\mathrm{1} \\ $$
Commented by 123456 last updated on 23/Mar/15
$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$${f}\left({x}−\mathrm{1}\right)={a}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}−\mathrm{1}\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} +\left({b}−\mathrm{2}{a}\right){x}+\left({a}−{b}+{c}\right) \\ $$$${f}\left({x}^{\mathrm{2}} \right)={ax}^{\mathrm{4}} +{bx}^{\mathrm{2}} +{c} \\ $$$$\begin{cases}{{a}={a}^{\mathrm{2}} }\\{\mathrm{0}=\mathrm{2}{a}\left({b}−{a}\right)}\\{{b}=\mathrm{2}{ac}+{b}^{\mathrm{2}} −\mathrm{3}{ab}+{a}^{\mathrm{2}} }\\{\mathrm{0}=\mathrm{2}\left({b}−{a}\right){c}−{b}^{\mathrm{2}} +{ab}}\\{{c}={c}\left({a}+{c}−{b}\right)}\end{cases} \\ $$$${f}\left({x}\right)=\mathrm{0}\vee{f}\left({x}\right)=\mathrm{1}\vee{f}\left({x}\right)={x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$
Commented by 123456 last updated on 23/Mar/15
$$\mathrm{suppose}\:\mathrm{that}\:{f}_{\mathrm{1}} \left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{and} \\ $$$${f}\left({x}\right)={f}_{\mathrm{1}} \left({x}\right)+{f}_{\mathrm{2}} \left({x}\right)\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{solution} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={f}\left({x}\right){f}\left({x}−\mathrm{1}\right) \\ $$$${f}_{\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{f}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)=\left[{f}_{\mathrm{1}} \left({x}\right)+{f}_{\mathrm{2}} \left({x}\right)\right]\left[{f}_{\mathrm{1}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}−\mathrm{1}\right)\right] \\ $$$$={f}_{\mathrm{1}} \left({x}\right){f}_{\mathrm{1}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{1}} \left({x}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}\right){f}_{\mathrm{1}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right) \\ $$$${f}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)={f}_{\mathrm{1}} \left({x}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}\right){f}_{\mathrm{1}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right) \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\mathrm{0}\Rightarrow{f}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)={f}_{\mathrm{2}} \left({x}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right) \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\mathrm{1}\Rightarrow{f}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)={f}_{\mathrm{2}} \left({x}−\mathrm{1}\right)+{f}_{\mathrm{2}} \left({x}\right)+{f}_{\mathrm{2}} \left({x}−\mathrm{1}\right){f}_{\mathrm{2}} \left({x}−\mathrm{1}\right) \\ $$
Commented by prakash jain last updated on 23/Mar/15
$$\mathrm{Thanks}.\:\mathrm{Then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{generalize}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{to} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{{n}} \:\mathrm{where}\:{n}\in\mathbb{N} \\ $$$$\mathrm{Since}\:\mathrm{raising}\:\mathrm{to}\:\mathrm{any}\:\mathrm{power}\:\mathrm{won}'\mathrm{t}\:\mathrm{make} \\ $$$$\mathrm{any}\:\mathrm{difference}\:\mathrm{in}\:\mathrm{equality}. \\ $$
Commented by prakash jain last updated on 23/Mar/15
$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+{x}−\mathrm{1}+\mathrm{1}\right) \\ $$$$=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$