Prove that every prime number divides some number in the sequence $ a_n = 2^n+3^n+6^n-1 $ [closed]
$begingroup$
Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:
$$ a_n = 2^n+3^n+6^n-1 $$
Show that every prime number divides some number in that sequence.
sequences-and-series elementary-number-theory prime-numbers recurrence-relations
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
$endgroup$
closed as off-topic by RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:
$$ a_n = 2^n+3^n+6^n-1 $$
Show that every prime number divides some number in that sequence.
sequences-and-series elementary-number-theory prime-numbers recurrence-relations
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
$endgroup$
closed as off-topic by RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
1
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
1
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08
add a comment |
$begingroup$
Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:
$$ a_n = 2^n+3^n+6^n-1 $$
Show that every prime number divides some number in that sequence.
sequences-and-series elementary-number-theory prime-numbers recurrence-relations
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
$endgroup$
Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:
$$ a_n = 2^n+3^n+6^n-1 $$
Show that every prime number divides some number in that sequence.
sequences-and-series elementary-number-theory prime-numbers recurrence-relations
sequences-and-series elementary-number-theory prime-numbers recurrence-relations
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
edited Jan 23 at 2:44
JimmyK4542
41.2k245107
41.2k245107
asked Jan 23 at 1:46
DoodDood
487
asked Jan 23 at 1:46
DoodDood
487
asked Jan 23 at 1:46
DoodDood
487
487
closed as off-topic by RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Jyrki Lahtonen, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
1
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
1
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08
add a comment |
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
1
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
1
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
1
1
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
1
1
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then
begin{align*}
6a_{p-2} & equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 pmod{p}\
& equiv 3cdot 2^{p-1}+2cdot 3^{p-1}+6^{p-1}-6pmod{p}\
& equiv 3+2+1 -6pmod{p}\
& equiv 0 pmod{p}.
end{align*}
For $p>3$, we will have $gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get
$$a_{p-2} equiv 0 pmod{p}$$
Now you can deal with $p=2,3$ as special case.
edited Jan 23 at 2:01
Robert Israel
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
$endgroup$
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then
begin{align*}
6a_{p-2} & equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 pmod{p}\
& equiv 3cdot 2^{p-1}+2cdot 3^{p-1}+6^{p-1}-6pmod{p}\
& equiv 3+2+1 -6pmod{p}\
& equiv 0 pmod{p}.
end{align*}
For $p>3$, we will have $gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get
$$a_{p-2} equiv 0 pmod{p}$$
Now you can deal with $p=2,3$ as special case.
edited Jan 23 at 2:01
Robert Israel
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
$endgroup$
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
add a comment |
$begingroup$
Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then
begin{align*}
6a_{p-2} & equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 pmod{p}\
& equiv 3cdot 2^{p-1}+2cdot 3^{p-1}+6^{p-1}-6pmod{p}\
& equiv 3+2+1 -6pmod{p}\
& equiv 0 pmod{p}.
end{align*}
For $p>3$, we will have $gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get
$$a_{p-2} equiv 0 pmod{p}$$
Now you can deal with $p=2,3$ as special case.
edited Jan 23 at 2:01
Robert Israel
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
$endgroup$
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
add a comment |
$begingroup$
Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then
begin{align*}
6a_{p-2} & equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 pmod{p}\
& equiv 3cdot 2^{p-1}+2cdot 3^{p-1}+6^{p-1}-6pmod{p}\
& equiv 3+2+1 -6pmod{p}\
& equiv 0 pmod{p}.
end{align*}
For $p>3$, we will have $gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get
$$a_{p-2} equiv 0 pmod{p}$$
Now you can deal with $p=2,3$ as special case.
edited Jan 23 at 2:01
Robert Israel
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
$endgroup$
Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then
begin{align*}
6a_{p-2} & equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 pmod{p}\
& equiv 3cdot 2^{p-1}+2cdot 3^{p-1}+6^{p-1}-6pmod{p}\
& equiv 3+2+1 -6pmod{p}\
& equiv 0 pmod{p}.
end{align*}
For $p>3$, we will have $gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get
$$a_{p-2} equiv 0 pmod{p}$$
Now you can deal with $p=2,3$ as special case.
edited Jan 23 at 2:01
Robert Israel
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
edited Jan 23 at 2:01
Robert Israel
326k23215469
edited Jan 23 at 2:01
Robert Israel
326k23215469
edited Jan 23 at 2:01
Robert Israel
326k23215469
326k23215469
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
answered Jan 23 at 2:00
Anurag AAnurag A
26.3k12251
26.3k12251
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
add a comment |
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
3
3
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
$begingroup$
Intuitively, taking $n=-1$ would give $a_{-1} = frac12 + frac13 + frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$.
$endgroup$
– Misha Lavrov
Jan 23 at 2:02
1
1
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
$begingroup$
Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response.
$endgroup$
– Dood
Jan 23 at 2:07
add a comment |
$begingroup$
I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks.
$endgroup$
– John Omielan
Jan 23 at 1:53
1
$begingroup$
This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed.
$endgroup$
– Aaron
Jan 23 at 15:07
1
$begingroup$
Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing.
$endgroup$
– Aaron
Jan 23 at 15:08