When is $n^{2015}+n+1$ prime?
$n in mathbb{N}$.
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
add a comment |
$n in mathbb{N}$.
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
your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16
add a comment |
$n in mathbb{N}$.
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
$n in mathbb{N}$.
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
edited Jan 4 at 21:58
greedoid
38.6k114797
38.6k114797
asked Aug 10 '17 at 15:14
AdienlAdienl
551416
551416
your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16
add a comment |
your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16
your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16
your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16
add a comment |
2 Answers
2
active
oldest
votes
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 $n=1$.
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
add a comment |
$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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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 $n=1$.
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
add a comment |
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 $n=1$.
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
add a comment |
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 $n=1$.
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 $n=1$.
answered Aug 10 '17 at 15:40
greedoidgreedoid
38.6k114797
38.6k114797
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
add a comment |
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
This would be a good contest-math Q at some level of competition.
– DanielWainfleet
Jan 5 at 1:05
add a comment |
$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
add a comment |
$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
add a comment |
$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...
$2015equiv 2pmod{3}$ implies that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...
answered Aug 10 '17 at 15:41
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
add a comment |
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
What is that function?
– Adienl
Aug 10 '17 at 15:54
What is that function?
– Adienl
Aug 10 '17 at 15:54
1
1
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
@Adienl: the standard notation for cyclotomic polynomials: en.wikipedia.org/wiki/Cyclotomic_polynomial
– Jack D'Aurizio
Aug 10 '17 at 15:57
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
Hi. I do not understand. Why does $2015equiv 2pmod{3}$ imply that $Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$?!
– stressed out
Jul 13 '18 at 8:02
1
1
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
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.
– Jack D'Aurizio
Jul 13 '18 at 9:07
add a comment |
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your term can be factorized!
– Dr. Sonnhard Graubner
Aug 10 '17 at 19:16