When is $n^{2015}+n+1$ prime?












14












$begingroup$


$n in mathbb{N}$.



I think this seems true only for $1$.



I tried to show that $4n-1,2n+1,2^{n+1}-1$ divides the given expression but didn't succeed. I have only thought about this through the way of modular arithmetic. Someone suggested to me to use the complex roots of this equation, but didn't specify how. Are there any other results I can use for this problem(I may not know them)?



Please provide only hints if you do solve it. Thank you.










share|cite|improve this question











$endgroup$












  • $begingroup$
    your term can be factorized!
    $endgroup$
    – Dr. Sonnhard Graubner
    Aug 10 '17 at 19:16
















14












$begingroup$


$n in mathbb{N}$.



I think this seems true only for $1$.



I tried to show that $4n-1,2n+1,2^{n+1}-1$ divides the given expression but didn't succeed. I have only thought about this through the way of modular arithmetic. Someone suggested to me to use the complex roots of this equation, but didn't specify how. Are there any other results I can use for this problem(I may not know them)?



Please provide only hints if you do solve it. Thank you.










share|cite|improve this question











$endgroup$












  • $begingroup$
    your term can be factorized!
    $endgroup$
    – Dr. Sonnhard Graubner
    Aug 10 '17 at 19:16














14












14








14


5



$begingroup$


$n in mathbb{N}$.



I think this seems true only for $1$.



I tried to show that $4n-1,2n+1,2^{n+1}-1$ divides the given expression but didn't succeed. I have only thought about this through the way of modular arithmetic. Someone suggested to me to use the complex roots of this equation, but didn't specify how. Are there any other results I can use for this problem(I may not know them)?



Please provide only hints if you do solve it. Thank you.










share|cite|improve this question











$endgroup$




$n in mathbb{N}$.



I think this seems true only for $1$.



I tried to show that $4n-1,2n+1,2^{n+1}-1$ divides the given expression but didn't succeed. I have only thought about this through the way of modular arithmetic. Someone suggested to me to use the complex roots of this equation, but didn't specify how. Are there any other results I can use for this problem(I may not know them)?



Please provide only hints if you do solve it. Thank you.







number-theory elementary-number-theory polynomials prime-numbers divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:58









greedoid

41k1149101




41k1149101










asked Aug 10 '17 at 15:14









AdienlAdienl

551417




551417












  • $begingroup$
    your term can be factorized!
    $endgroup$
    – Dr. Sonnhard Graubner
    Aug 10 '17 at 19:16


















  • $begingroup$
    your term can be factorized!
    $endgroup$
    – Dr. Sonnhard Graubner
    Aug 10 '17 at 19:16
















$begingroup$
your term can be factorized!
$endgroup$
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16




$begingroup$
your term can be factorized!
$endgroup$
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16










2 Answers
2






active

oldest

votes


















27












$begingroup$

begin{eqnarray*}
n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \
&=& n^2(n^{2013} -1) + n^2+n+1 \
&=& n^2(n^3-1)underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \
&=& (n^2+n+1)(an^2(n-1)+1) \
&=&
end{eqnarray*}
Since $n^2+n+1geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would be a good contest-math Q at some level of competition.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 1:05



















11












$begingroup$

$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is that function?
    $endgroup$
    – Adienl
    Aug 10 '17 at 15:54






  • 1




    $begingroup$
    @Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
    $endgroup$
    – Jack D'Aurizio
    Aug 10 '17 at 15:57










  • $begingroup$
    Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
    $endgroup$
    – stressed out
    Jul 13 '18 at 8:02






  • 1




    $begingroup$
    Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '18 at 9:07











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%2f2389232%2fwhen-is-n2015n1-prime%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









27












$begingroup$

begin{eqnarray*}
n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \
&=& n^2(n^{2013} -1) + n^2+n+1 \
&=& n^2(n^3-1)underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \
&=& (n^2+n+1)(an^2(n-1)+1) \
&=&
end{eqnarray*}
Since $n^2+n+1geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would be a good contest-math Q at some level of competition.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 1:05
















27












$begingroup$

begin{eqnarray*}
n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \
&=& n^2(n^{2013} -1) + n^2+n+1 \
&=& n^2(n^3-1)underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \
&=& (n^2+n+1)(an^2(n-1)+1) \
&=&
end{eqnarray*}
Since $n^2+n+1geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This would be a good contest-math Q at some level of competition.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 1:05














27












27








27





$begingroup$

begin{eqnarray*}
n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \
&=& n^2(n^{2013} -1) + n^2+n+1 \
&=& n^2(n^3-1)underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \
&=& (n^2+n+1)(an^2(n-1)+1) \
&=&
end{eqnarray*}
Since $n^2+n+1geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.






share|cite|improve this answer









$endgroup$



begin{eqnarray*}
n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \
&=& n^2(n^{2013} -1) + n^2+n+1 \
&=& n^2(n^3-1)underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \
&=& (n^2+n+1)(an^2(n-1)+1) \
&=&
end{eqnarray*}
Since $n^2+n+1geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 10 '17 at 15:40









greedoidgreedoid

41k1149101




41k1149101












  • $begingroup$
    This would be a good contest-math Q at some level of competition.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 1:05


















  • $begingroup$
    This would be a good contest-math Q at some level of competition.
    $endgroup$
    – DanielWainfleet
    Jan 5 at 1:05
















$begingroup$
This would be a good contest-math Q at some level of competition.
$endgroup$
– DanielWainfleet
Jan 5 at 1:05




$begingroup$
This would be a good contest-math Q at some level of competition.
$endgroup$
– DanielWainfleet
Jan 5 at 1:05











11












$begingroup$

$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is that function?
    $endgroup$
    – Adienl
    Aug 10 '17 at 15:54






  • 1




    $begingroup$
    @Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
    $endgroup$
    – Jack D'Aurizio
    Aug 10 '17 at 15:57










  • $begingroup$
    Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
    $endgroup$
    – stressed out
    Jul 13 '18 at 8:02






  • 1




    $begingroup$
    Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '18 at 9:07
















11












$begingroup$

$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What is that function?
    $endgroup$
    – Adienl
    Aug 10 '17 at 15:54






  • 1




    $begingroup$
    @Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
    $endgroup$
    – Jack D'Aurizio
    Aug 10 '17 at 15:57










  • $begingroup$
    Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
    $endgroup$
    – stressed out
    Jul 13 '18 at 8:02






  • 1




    $begingroup$
    Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '18 at 9:07














11












11








11





$begingroup$

$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...






share|cite|improve this answer









$endgroup$



$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 10 '17 at 15:41









Jack D'AurizioJack D'Aurizio

289k33281661




289k33281661












  • $begingroup$
    What is that function?
    $endgroup$
    – Adienl
    Aug 10 '17 at 15:54






  • 1




    $begingroup$
    @Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
    $endgroup$
    – Jack D'Aurizio
    Aug 10 '17 at 15:57










  • $begingroup$
    Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
    $endgroup$
    – stressed out
    Jul 13 '18 at 8:02






  • 1




    $begingroup$
    Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '18 at 9:07


















  • $begingroup$
    What is that function?
    $endgroup$
    – Adienl
    Aug 10 '17 at 15:54






  • 1




    $begingroup$
    @Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
    $endgroup$
    – Jack D'Aurizio
    Aug 10 '17 at 15:57










  • $begingroup$
    Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
    $endgroup$
    – stressed out
    Jul 13 '18 at 8:02






  • 1




    $begingroup$
    Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
    $endgroup$
    – Jack D'Aurizio
    Jul 13 '18 at 9:07
















$begingroup$
What is that function?
$endgroup$
– Adienl
Aug 10 '17 at 15:54




$begingroup$
What is that function?
$endgroup$
– Adienl
Aug 10 '17 at 15:54




1




1




$begingroup$
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– Jack D'Aurizio
Aug 10 '17 at 15:57




$begingroup$
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– Jack D'Aurizio
Aug 10 '17 at 15:57












$begingroup$
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
$endgroup$
– stressed out
Jul 13 '18 at 8:02




$begingroup$
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
$endgroup$
– stressed out
Jul 13 '18 at 8:02




1




1




$begingroup$
Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
$endgroup$
– Jack D'Aurizio
Jul 13 '18 at 9:07




$begingroup$
Because n raised to the 2015th power is congruent to n^2 mod n^3-1, hence n^2015 is congruent to n^2 mod n^2+n+1 too, since the last polynomial is a divisor of n^3-1.
$endgroup$
– Jack D'Aurizio
Jul 13 '18 at 9:07


















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%2f2389232%2fwhen-is-n2015n1-prime%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

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅