Menu Close

Find-all-functions-h-Z-Z-such-that-h-x-y-h-xy-h-x-h-y-1-for-all-x-y-Z-

Tinku Tara 0 Comments
Pin




Question Number 4751 by Yozzii last updated on 04/Mar/16
Find all functions h:Z→Z such that h(x+y)+h(xy)=h(x)h(y)+1 for all x,y∈Z.
$${Find}\:{all}\:{functions}\:{h}:\mathbb{Z}\rightarrow\mathbb{Z}\:{such}\:{that} \\ $$$${h}\left({x}+{y}\right)+{h}\left({xy}\right)={h}\left({x}\right){h}\left({y}\right)+\mathrm{1} \\ $$$${for}\:{all}\:{x},{y}\in\mathbb{Z}. \\ $$
Commented by prakash jain last updated on 06/Mar/16
y=0 h(x)+h(0)=h(x)h(0)+1 h(x)(1−h(0))=1−h(0) ⇒h(x)=1 or h(0)=1 One solution h(x)=1 case 2: h(0)=1 y=−x h(0)+h(−x^2 )=[h(x)][h(−x)]+1 h(x)h(−x)=h(−x^2 )
$${y}=\mathrm{0} \\ $$$${h}\left({x}\right)+{h}\left(\mathrm{0}\right)={h}\left({x}\right){h}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$${h}\left({x}\right)\left(\mathrm{1}−{h}\left(\mathrm{0}\right)\right)=\mathrm{1}−{h}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{h}\left({x}\right)=\mathrm{1}\:{or}\:{h}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{One}\:\mathrm{solution}\:{h}\left({x}\right)=\mathrm{1} \\ $$$${case}\:\mathrm{2}:\:{h}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${y}=−{x} \\ $$$${h}\left(\mathrm{0}\right)+{h}\left(−{x}^{\mathrm{2}} \right)=\left[{h}\left({x}\right)\right]\left[{h}\left(−{x}\right)\right]+\mathrm{1} \\ $$$${h}\left({x}\right){h}\left(−{x}\right)={h}\left(−{x}^{\mathrm{2}} \right) \\ $$
Commented by prakash jain last updated on 07/Mar/16
Since only integers are required h(0)=1 x=1,y=−1 h(0)+h(−1)=h(1)h(−1)+1⇒h(1)=1 or h(−1)=0 case h(1)=1 h(n+1)+h(n)=h(n)h(1)+1 h(n+1)=1 ⇒h(x)=1 for x≥0 ...(A) Let h(−1)=k x=−2,y=1 h(−1)+h(−2)=h(−1)h(1)+1 h(−2)=1 x=−1,y=−1 h(−2)+h(1)=h(−1)h(1)+1 h(−2)=k x=1,y=−3 h(−2)+h(−3)=h(−3)h(1)+1 h(−2)=1⇒k=1 x=−2,y=−1 h(−3)+h(2)=h(−2)h(1)+1⇒h(−3)=1 if h(1)=1⇒h(x)=1 ∀x∈Z case h(−1)=0 Corrections in blue. x=2,y=−1 h(1)+h(−2)=h(2)h(−1)+1 x=−n,y=−1 h(−n−1)+h(−n)=h(−n)h(−1)+1 h(−n−1)=−h(−n) The above line also wrong h(−1)=0 h(−n−1)=1−h(−n) ⇒h(1)=1 (wrong) ⇒h(x)=1 contradicts h(−1)=0 too many mistakes in handling h(−1)=0 will redo. So it looks like h(x)=1 is the only solution. Are there any other solutions in answer?
$$\mathrm{Since}\:\mathrm{only}\:\mathrm{integers}\:\mathrm{are}\:\mathrm{required}\: \\ $$$${h}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${x}=\mathrm{1},{y}=−\mathrm{1} \\ $$$${h}\left(\mathrm{0}\right)+{h}\left(−\mathrm{1}\right)={h}\left(\mathrm{1}\right){h}\left(−\mathrm{1}\right)+\mathrm{1}\Rightarrow{h}\left(\mathrm{1}\right)=\mathrm{1}\:{or}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$${case}\:{h}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:{h}\left({n}+\mathrm{1}\right)+{h}\left({n}\right)={h}\left({n}\right){h}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:{h}\left({n}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\Rightarrow{h}\left({x}\right)=\mathrm{1}\:{for}\:{x}\geqslant\mathrm{0}\:\:\:…\left({A}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{Let}\:{h}\left(−\mathrm{1}\right)={k} \\ $$$$\:\:\:\:\:\:\:{x}=−\mathrm{2},{y}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{h}\left(−\mathrm{1}\right)+{h}\left(−\mathrm{2}\right)={h}\left(−\mathrm{1}\right){h}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{h}\left(−\mathrm{2}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{x}=−\mathrm{1},{y}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}\left(−\mathrm{2}\right)+{h}\left(\mathrm{1}\right)={h}\left(−\mathrm{1}\right){h}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}\left(−\mathrm{2}\right)={k} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{1},{y}=−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}\left(−\mathrm{2}\right)+{h}\left(−\mathrm{3}\right)={h}\left(−\mathrm{3}\right){h}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}\left(−\mathrm{2}\right)=\mathrm{1}\Rightarrow{k}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{x}=−\mathrm{2},{y}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{h}\left(−\mathrm{3}\right)+{h}\left(\mathrm{2}\right)={h}\left(−\mathrm{2}\right){h}\left(\mathrm{1}\right)+\mathrm{1}\Rightarrow{h}\left(−\mathrm{3}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{if}\:{h}\left(\mathrm{1}\right)=\mathrm{1}\Rightarrow{h}\left({x}\right)=\mathrm{1}\:\forall{x}\in\mathbb{Z} \\ $$$${case}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{Corrections}\:\mathrm{in}\:\mathrm{blue}. \\ $$$$\:\:\:\:\:{x}=\mathrm{2},{y}=−\mathrm{1} \\ $$$$\:\:\:\:\:{h}\left(\mathrm{1}\right)+{h}\left(−\mathrm{2}\right)={h}\left(\mathrm{2}\right){h}\left(−\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:{x}=−{n},{y}=−\mathrm{1} \\ $$$$\:\:\:\:\:{h}\left(−{n}−\mathrm{1}\right)+{h}\left(−{n}\right)={h}\left(−{n}\right){h}\left(−\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:{h}\left(−{n}−\mathrm{1}\right)=−{h}\left(−{n}\right) \\ $$$$\:\:\:\:\:\mathrm{The}\:\mathrm{above}\:\mathrm{line}\:\mathrm{also}\:\mathrm{wrong}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{h}\left(−{n}−\mathrm{1}\right)=\mathrm{1}−{h}\left(−{n}\right) \\ $$$$\:\:\:\:\:\Rightarrow{h}\left(\mathrm{1}\right)=\mathrm{1}\:\left({wrong}\right) \\ $$$$\:\:\:\:\:\Rightarrow{h}\left({x}\right)=\mathrm{1}\:{contradicts}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{too}\:{many}\:{mistakes}\:{in}\:{handling}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{will}\:{redo}. \\ $$$$\mathrm{So}\:\mathrm{it}\:\mathrm{looks}\:\mathrm{like}\:{h}\left({x}\right)=\mathrm{1}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}. \\ $$$$\mathrm{Are}\:\mathrm{there}\:\mathrm{any}\:\mathrm{other}\:\mathrm{solutions}\:\mathrm{in}\:\mathrm{answer}? \\ $$
Commented by Yozzii last updated on 07/Mar/16
(1) h(n)=1 (n∈Z) (2) h(2n)=1;h(2n+1)=0 (n∈Z) (3) h(n)=n+1 (n∈Z) I′m trying out national olympiad questions (year:1999) from a QandA book. So, I′m sharing those which I hope to try with some degree of success.
$$\left(\mathrm{1}\right)\:{h}\left({n}\right)=\mathrm{1}\:\left({n}\in\mathbb{Z}\right) \\ $$$$\left(\mathrm{2}\right)\:{h}\left(\mathrm{2}{n}\right)=\mathrm{1};{h}\left(\mathrm{2}{n}+\mathrm{1}\right)=\mathrm{0}\:\left({n}\in\mathbb{Z}\right) \\ $$$$\left(\mathrm{3}\right)\:{h}\left({n}\right)={n}+\mathrm{1}\:\:\left({n}\in\mathbb{Z}\right) \\ $$$${I}'{m}\:{trying}\:{out}\:{national}\:{olympiad} \\ $$$${questions}\:\left({year}:\mathrm{1999}\right)\:{from}\:{a}\:{QandA}\: \\ $$$${book}.\:{So},\:{I}'{m}\:{sharing}\:{those}\:{which} \\ $$$${I}\:{hope}\:{to}\:{try}\:{with}\:{some}\:{degree}\:{of}\: \\ $$$${success}. \\ $$
Commented by prakash jain last updated on 07/Mar/16
I will check for mistakes in my arguments. I thin mistake may be in handling the case h(−1)=0
$$\mathrm{I}\:\mathrm{will}\:\mathrm{check}\:\mathrm{for}\:\mathrm{mistakes}\:\mathrm{in}\:\mathrm{my}\:\mathrm{arguments}. \\ $$$$\mathrm{I}\:\mathrm{thin}\:\mathrm{mistake}\:\mathrm{may}\:\mathrm{be}\:\mathrm{in}\:\mathrm{handling}\:\mathrm{the}\:\mathrm{case} \\ $$$${h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by prakash jain last updated on 07/Mar/16
h(0)=1 case h(−1)=0 h(x+y)+h(xy)=h(x)h(y)+1 x=n,y=−1 h(n−1)+h(−n)=h(n)h(−1)+1 h(n−1)=1−h(−n) The above is linear equation solution a: h(n)=an+b a(n−1)+b=1−(−an+b) an−a+b=1+an−b −a+b=1−b⇒b=((1+a)/2) h(n)=an+((1+a)/2) h(−1)=0⇒0=−a+((1+a)/2)⇒a=1 h(n)=n+1 (solution a) h(x+y)+h(xy)=h(x)h(y)+1 x+y+1+(xy+1)=(x+1)(y+1)+1 h(−n−1)=1−h(−n) or h(n−1)=1−h(n) given h(−1)=0,h(0)=1 h(n)=1−h(n−1) h(1)=1−h(0)=0 h(2)=1−h(1)=1 or h(n)= { ((0 n odd)),((1 n even)) :}
$${h}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{case}\:{h}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$${h}\left({x}+{y}\right)+{h}\left({xy}\right)={h}\left({x}\right){h}\left({y}\right)+\mathrm{1} \\ $$$${x}={n},{y}=−\mathrm{1} \\ $$$${h}\left({n}−\mathrm{1}\right)+{h}\left(−{n}\right)={h}\left({n}\right){h}\left(−\mathrm{1}\right)+\mathrm{1} \\ $$$${h}\left({n}−\mathrm{1}\right)=\mathrm{1}−{h}\left(−{n}\right) \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{is}\:\mathrm{linear}\:\mathrm{equation} \\ $$$${solution}\:{a}: \\ $$$${h}\left({n}\right)={an}+{b} \\ $$$${a}\left({n}−\mathrm{1}\right)+{b}=\mathrm{1}−\left(−{an}+{b}\right) \\ $$$${an}−{a}+{b}=\mathrm{1}+{an}−{b} \\ $$$$−{a}+{b}=\mathrm{1}−{b}\Rightarrow{b}=\frac{\mathrm{1}+{a}}{\mathrm{2}} \\ $$$${h}\left({n}\right)={an}+\frac{\mathrm{1}+{a}}{\mathrm{2}} \\ $$$${h}\left(−\mathrm{1}\right)=\mathrm{0}\Rightarrow\mathrm{0}=−{a}+\frac{\mathrm{1}+{a}}{\mathrm{2}}\Rightarrow{a}=\mathrm{1} \\ $$$${h}\left({n}\right)={n}+\mathrm{1}\:\left({solution}\:{a}\right) \\ $$$${h}\left({x}+{y}\right)+{h}\left({xy}\right)={h}\left({x}\right){h}\left({y}\right)+\mathrm{1} \\ $$$${x}+{y}+\mathrm{1}+\left({xy}+\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)+\mathrm{1} \\ $$$$ \\ $$$${h}\left(−{n}−\mathrm{1}\right)=\mathrm{1}−{h}\left(−{n}\right) \\ $$$$\mathrm{or}\:{h}\left({n}−\mathrm{1}\right)=\mathrm{1}−{h}\left({n}\right)\:\:\mathrm{given}\:{h}\left(−\mathrm{1}\right)=\mathrm{0},{h}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${h}\left({n}\right)=\mathrm{1}−{h}\left({n}−\mathrm{1}\right) \\ $$$${h}\left(\mathrm{1}\right)=\mathrm{1}−{h}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${h}\left(\mathrm{2}\right)=\mathrm{1}−{h}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${or}\:{h}\left({n}\right)=\begin{cases}{\mathrm{0}\:{n}\:{odd}}\\{\mathrm{1}\:{n}\:{even}}\end{cases} \\ $$
Commented by Yozzii last updated on 07/Mar/16
Looks like thats one approach to figuring out the answer. Nice!
$${Looks}\:{like}\:{thats}\:{one}\:{approach}\:{to}\: \\ $$$${figuring}\:{out}\:{the}\:{answer}.\:{Nice}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *