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.






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







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.






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







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.






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







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




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











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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


















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