Proof verification for polynomial solutions to $xP(x-1)=(x-2)P(x)$












2












$begingroup$


Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.




Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$




Here is my solution, is it right?



Replace $x$ with $0$ and $1$ and clearly, they equal $0$.



Suppose there is another positive root $rneq 1$.



Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.



Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.



Now suppose there is another negative root $c$.



Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.



Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.



So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.






number-theory proof-verification polynomials roots functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why is $P(r+1)=0$?
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:36










  • $begingroup$
    It is better if you go by carat foot steps...
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:37








  • 2




    $begingroup$
    What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 12:59


















2












$begingroup$


Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.




Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$




Here is my solution, is it right?



Replace $x$ with $0$ and $1$ and clearly, they equal $0$.



Suppose there is another positive root $rneq 1$.



Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.



Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.



Now suppose there is another negative root $c$.



Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.



Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.



So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.






number-theory proof-verification polynomials roots functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why is $P(r+1)=0$?
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:36










  • $begingroup$
    It is better if you go by carat foot steps...
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:37








  • 2




    $begingroup$
    What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 12:59
















2












2








2





$begingroup$


Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.




Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$




Here is my solution, is it right?



Replace $x$ with $0$ and $1$ and clearly, they equal $0$.



Suppose there is another positive root $rneq 1$.



Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.



Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.



Now suppose there is another negative root $c$.



Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.



Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.



So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.






number-theory proof-verification polynomials roots functional-equations







share|cite|improve this question











$endgroup$




Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.




Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$




Here is my solution, is it right?



Replace $x$ with $0$ and $1$ and clearly, they equal $0$.



Suppose there is another positive root $rneq 1$.



Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.



Thus if there exists another positive root $rneq 1$, then we have an infinite number of roots which means $P(x)=0$.



Now suppose there is another negative root $c$.



Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.



Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.



So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.






number-theory proof-verification polynomials roots functional-equations




number-theory proof-verification polynomials roots functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 12:45









Maria Mazur

50.7k1362126




50.7k1362126










asked Dec 31 '18 at 12:28









user627514user627514

393




393












  • $begingroup$
    Why is $P(r+1)=0$?
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:36










  • $begingroup$
    It is better if you go by carat foot steps...
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:37








  • 2




    $begingroup$
    What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 12:59




















  • $begingroup$
    Why is $P(r+1)=0$?
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:36










  • $begingroup$
    It is better if you go by carat foot steps...
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 12:37








  • 2




    $begingroup$
    What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
    $endgroup$
    – A. Pongrácz
    Dec 31 '18 at 12:59


















$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36




$begingroup$
Why is $P(r+1)=0$?
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:36












$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37






$begingroup$
It is better if you go by carat foot steps...
$endgroup$
– Maria Mazur
Dec 31 '18 at 12:37






2




2




$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59






$begingroup$
What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example.
$endgroup$
– A. Pongrácz
Dec 31 '18 at 12:59












1 Answer
1






active

oldest

votes


















1












$begingroup$

If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).



So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Was about to upvote when I realsied that you're not actually answering the question.
    $endgroup$
    – Git Gud
    Dec 31 '18 at 17:29










  • $begingroup$
    Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 18:03














Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057661%2fproof-verification-for-polynomial-solutions-to-xpx-1-x-2px%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).



So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Was about to upvote when I realsied that you're not actually answering the question.
    $endgroup$
    – Git Gud
    Dec 31 '18 at 17:29










  • $begingroup$
    Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 18:03


















1












$begingroup$

If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).



So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Was about to upvote when I realsied that you're not actually answering the question.
    $endgroup$
    – Git Gud
    Dec 31 '18 at 17:29










  • $begingroup$
    Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 18:03
















1












1








1





$begingroup$

If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).



So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.






share|cite|improve this answer











$endgroup$



If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $xne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).



So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 13:03

























answered Dec 31 '18 at 12:44









Maria MazurMaria Mazur

50.7k1362126




50.7k1362126












  • $begingroup$
    Was about to upvote when I realsied that you're not actually answering the question.
    $endgroup$
    – Git Gud
    Dec 31 '18 at 17:29










  • $begingroup$
    Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 18:03




















  • $begingroup$
    Was about to upvote when I realsied that you're not actually answering the question.
    $endgroup$
    – Git Gud
    Dec 31 '18 at 17:29










  • $begingroup$
    Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
    $endgroup$
    – Maria Mazur
    Dec 31 '18 at 18:03


















$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29




$begingroup$
Was about to upvote when I realsied that you're not actually answering the question.
$endgroup$
– Git Gud
Dec 31 '18 at 17:29












$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03






$begingroup$
Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud
$endgroup$
– Maria Mazur
Dec 31 '18 at 18:03




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057661%2fproof-verification-for-polynomial-solutions-to-xpx-1-x-2px%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
Read more