A problem of factoring a polynomial with a hint
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
asked Feb 3 at 3:53
zimbra314zimbra314
628312
$endgroup$
add a comment |
$begingroup$
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
asked Feb 3 at 3:53
zimbra314zimbra314
628312
$endgroup$
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01
add a comment |
$begingroup$
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
asked Feb 3 at 3:53
zimbra314zimbra314
628312
$endgroup$
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
asked Feb 3 at 3:53
zimbra314zimbra314
628312
asked Feb 3 at 3:53
zimbra314zimbra314
628312
edited Feb 3 at 7:36
zimbra314
asked Feb 3 at 3:53
zimbra314zimbra314
628312
asked Feb 3 at 3:53
zimbra314zimbra314
628312
asked Feb 3 at 3:53
zimbra314zimbra314
628312
628312
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01
add a comment |
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01
1
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
add a comment |
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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.
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
add a comment |
$begingroup$
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.
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
add a comment |
$begingroup$
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.
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
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.
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
edited Feb 3 at 4:40
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
answered Feb 3 at 4:25
Eric WofseyEric Wofsey
187k14215344
187k14215344
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
add a comment |
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
$begingroup$
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
Feb 3 at 7:33
$begingroup$
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
Feb 3 at 7:33
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
$begingroup$
More precisely, it is the square of an irreducible polynomial (since otherwise it would have a nontrivial factorization with $Pneq Q$).
$endgroup$
– Eric Wofsey
Feb 3 at 7:52
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
What if $n=2$, $a_1 = 1$ and $a_2 = -1$?
$endgroup$
– darij grinberg
Feb 3 at 4:01