Prove that every prime number divides some number in the sequence $ a_n = 2^n+3^n+6^n-1 $ [closed]












0












$begingroup$


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







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$endgroup$



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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$



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














0












0








0


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 2:44









JimmyK4542

41.2k245107




41.2k245107










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















8












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
















8












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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














8












8








8





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 23 at 2:01









Robert Israel

326k23215469




326k23215469










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














  • 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



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