A problem of factoring a polynomial with a hint












1












$begingroup$


PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.



This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$



This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as




if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$




How does one solve this problem this hint?



Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7






polynomials contest-math irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago
















1












$begingroup$


PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.



This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$



This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as




if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$




How does one solve this problem this hint?



Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7






polynomials contest-math irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago














1












1








1





$begingroup$


PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.



This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$



This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as




if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$




How does one solve this problem this hint?



Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7






polynomials contest-math irreducible-polynomials







share|cite|improve this question











$endgroup$




PT $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ can not be factored into two smaller polynomial $P(x)$ and $Q(x)$ with integer coefficients, where $a_{i}$'s are all different integers.



This problem can be solved by considering the root of equation $P(x)Q(x)-1=0$



This problem comes from Terry Tao's book Solving mathematical problem(page 47), in which he gives a hint as




if P(x) and Q(x) are such factors then what can you say about $P(x)-Q(x)$




How does one solve this problem this hint?



Edit:
This appears not to be true as pointed out by Darji and Eric. For interested readers, The actual problem can be found here, page 47 Excercise 3.7






polynomials contest-math irreducible-polynomials




polynomials contest-math irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 mins ago







zimbra314

















asked 3 hours ago









zimbra314zimbra314

523212




523212








  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago














  • 1




    $begingroup$
    What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
    $endgroup$
    – darij grinberg
    3 hours ago








1




1




$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago




$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
3 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.



But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.



The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
    $endgroup$
    – zimbra314
    10 mins ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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%2f3098151%2fa-problem-of-factoring-a-polynomial-with-a-hint%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









4












$begingroup$

Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.



But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.



The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
    $endgroup$
    – zimbra314
    10 mins ago
















4












$begingroup$

Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.



But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.



The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
    $endgroup$
    – zimbra314
    10 mins ago














4












4








4





$begingroup$

Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.



But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.



The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.






share|cite|improve this answer











$endgroup$



Notice that $P(a_i)Q(a_i)=1$ for all $i$. But since $P$ and $Q$ have integer coefficients and the $a_i$ are integers, this means $P(a_i)=Q(a_i)=1$ or $P(a_i)=Q(a_i)=-1$ for each $i$. In particular, $a_i$ is a root of $P-Q$ for each $i$, so $P-Q$ is either $0$ or has degree at least $n$.



But, since $deg(PQ)=n$, the only way $P-Q$ can have degree at least $n$ is if one of $P$ and $Q$ has degree $n$ and the other is a constant. Presumably this possibility is meant to be excluded by the requirement that $P$ and $Q$ are "smaller" polynomials.



The only remaining possibility is that $P-Q=0$, so $P=Q$. This actually is possible--for instance, as darij grinberg commented, you could have $n=2$, $a_1=1$, and $a_2=-1$ and so the polynomial is $(x-1)(x+1)+1=x^2$ and so we can have $P(x)=Q(x)=x$. Another example (with $n=4$) is $$(x-1)(x-2)(x-3)(x-4)+1=(x^2-5x+5)^2.$$ So, the problem statement is not quite correct without some additional assumption to rule out this case.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 3 hours ago









Eric WofseyEric Wofsey

184k13213339




184k13213339












  • $begingroup$
    Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
    $endgroup$
    – zimbra314
    10 mins ago


















  • $begingroup$
    Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
    $endgroup$
    – zimbra314
    10 mins ago
















$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
10 mins ago




$begingroup$
Great example. So it appears if $(x-a_{1})(x-a_{2})..(x-a_{n})+1$ reducible then it is also a square of a polynomial?
$endgroup$
– zimbra314
10 mins ago


















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%2f3098151%2fa-problem-of-factoring-a-polynomial-with-a-hint%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

How to label and detect the document text images

Image
1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
Read more

Vallis Paradisi

Image
Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
Read more

Tabula Rosettana

Image
Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
Read more