Prove that every prime number divides some number in the sequence












1












$begingroup$


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.






linear-algebra sequences-and-series prime-numbers recurrence-relations







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  • $begingroup$
    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
    40 mins ago
















1












$begingroup$


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.






linear-algebra sequences-and-series prime-numbers recurrence-relations







share|cite









$endgroup$












  • $begingroup$
    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
    40 mins ago














1












1








1


1



$begingroup$


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.






linear-algebra sequences-and-series prime-numbers recurrence-relations







share|cite









$endgroup$




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.






linear-algebra sequences-and-series prime-numbers recurrence-relations




linear-algebra sequences-and-series prime-numbers recurrence-relations






share|cite













share|cite











share|cite




share|cite










asked 48 mins ago









DoodDood

357




357












  • $begingroup$
    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
    40 mins ago


















  • $begingroup$
    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
    40 mins ago
















$begingroup$
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
40 mins ago




$begingroup$
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
40 mins ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

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
    31 mins ago






  • 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
    27 mins ago











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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
    31 mins ago






  • 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
    27 mins ago
















6












$begingroup$

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
    31 mins ago






  • 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
    27 mins ago














6












6








6





$begingroup$

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 32 mins ago









Robert Israel

321k23210462




321k23210462










answered 33 mins ago









Anurag AAnurag A

25.9k12249




25.9k12249








  • 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
    31 mins ago






  • 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
    27 mins ago














  • 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
    31 mins ago






  • 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
    27 mins ago








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
31 mins ago




$begingroup$
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
31 mins ago




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
27 mins ago




$begingroup$
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
27 mins ago


















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