Every odd divisor is $equiv 1bmod 3$












5












$begingroup$


If $n$ is a positive integer with $nequiv 2pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1pmod 3$.



$$$$



I have done the following:



Let $d$ be an odd divisor of $n^2+n+1$.



Then $$n^2+n+1equiv 0bmod d Rightarrow (n-1)(n^2+n+1)equiv 0bmod d Rightarrow n^3-1equiv 0 bmod d Rightarrow n^3equiv 1bmod d$$



Since $nequiv 2pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$



We have that $n^3equiv 1bmod d$ therefore we get $$27k^3+54k^2+36k+8equiv 1bmod d Rightarrow 27k^3+54k^2+36k+7equiv 0bmod d$$



Is everything correct so far? How could we continue?






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 20:57
















5












$begingroup$


If $n$ is a positive integer with $nequiv 2pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1pmod 3$.



$$$$



I have done the following:



Let $d$ be an odd divisor of $n^2+n+1$.



Then $$n^2+n+1equiv 0bmod d Rightarrow (n-1)(n^2+n+1)equiv 0bmod d Rightarrow n^3-1equiv 0 bmod d Rightarrow n^3equiv 1bmod d$$



Since $nequiv 2pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$



We have that $n^3equiv 1bmod d$ therefore we get $$27k^3+54k^2+36k+8equiv 1bmod d Rightarrow 27k^3+54k^2+36k+7equiv 0bmod d$$



Is everything correct so far? How could we continue?






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 20:57














5












5








5


1



$begingroup$


If $n$ is a positive integer with $nequiv 2pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1pmod 3$.



$$$$



I have done the following:



Let $d$ be an odd divisor of $n^2+n+1$.



Then $$n^2+n+1equiv 0bmod d Rightarrow (n-1)(n^2+n+1)equiv 0bmod d Rightarrow n^3-1equiv 0 bmod d Rightarrow n^3equiv 1bmod d$$



Since $nequiv 2pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$



We have that $n^3equiv 1bmod d$ therefore we get $$27k^3+54k^2+36k+8equiv 1bmod d Rightarrow 27k^3+54k^2+36k+7equiv 0bmod d$$



Is everything correct so far? How could we continue?






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




If $n$ is a positive integer with $nequiv 2pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1pmod 3$.



$$$$



I have done the following:



Let $d$ be an odd divisor of $n^2+n+1$.



Then $$n^2+n+1equiv 0bmod d Rightarrow (n-1)(n^2+n+1)equiv 0bmod d Rightarrow n^3-1equiv 0 bmod d Rightarrow n^3equiv 1bmod d$$



Since $nequiv 2pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$



We have that $n^3equiv 1bmod d$ therefore we get $$27k^3+54k^2+36k+8equiv 1bmod d Rightarrow 27k^3+54k^2+36k+7equiv 0bmod d$$



Is everything correct so far? How could we continue?






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 20:56









Mary StarMary Star

3,10282473




3,10282473








  • 1




    $begingroup$
    Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 20:57














  • 1




    $begingroup$
    Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 20:57








1




1




$begingroup$
Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
$endgroup$
– Mindlack
Dec 15 '18 at 20:57




$begingroup$
Try for prime divisors and consider the multiplicative order of $n$ mod $p$. Note that $n^2+n+1$ is odd.
$endgroup$
– Mindlack
Dec 15 '18 at 20:57










1 Answer
1






active

oldest

votes


















6












$begingroup$

Hint: Start with odd prime divisors. As you have correctly shown, for any $pmid n^2+n+1$ we have that
$$n^3equiv 1pmod p.$$
Hence $text{ord}_{p}(n)mid 3$. Since $nequiv 2notequiv 1pmod p$ we know that $text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $text{ord}_m(x)mid phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:23






  • 2




    $begingroup$
    @MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:24












  • $begingroup$
    It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:51






  • 2




    $begingroup$
    @MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:53










  • $begingroup$
    Great!! Thanks a lot!! :-)
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:59











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









6












$begingroup$

Hint: Start with odd prime divisors. As you have correctly shown, for any $pmid n^2+n+1$ we have that
$$n^3equiv 1pmod p.$$
Hence $text{ord}_{p}(n)mid 3$. Since $nequiv 2notequiv 1pmod p$ we know that $text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $text{ord}_m(x)mid phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:23






  • 2




    $begingroup$
    @MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:24












  • $begingroup$
    It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:51






  • 2




    $begingroup$
    @MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:53










  • $begingroup$
    Great!! Thanks a lot!! :-)
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:59
















6












$begingroup$

Hint: Start with odd prime divisors. As you have correctly shown, for any $pmid n^2+n+1$ we have that
$$n^3equiv 1pmod p.$$
Hence $text{ord}_{p}(n)mid 3$. Since $nequiv 2notequiv 1pmod p$ we know that $text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $text{ord}_m(x)mid phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:23






  • 2




    $begingroup$
    @MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:24












  • $begingroup$
    It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:51






  • 2




    $begingroup$
    @MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:53










  • $begingroup$
    Great!! Thanks a lot!! :-)
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:59














6












6








6





$begingroup$

Hint: Start with odd prime divisors. As you have correctly shown, for any $pmid n^2+n+1$ we have that
$$n^3equiv 1pmod p.$$
Hence $text{ord}_{p}(n)mid 3$. Since $nequiv 2notequiv 1pmod p$ we know that $text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $text{ord}_m(x)mid phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?






share|cite|improve this answer









$endgroup$



Hint: Start with odd prime divisors. As you have correctly shown, for any $pmid n^2+n+1$ we have that
$$n^3equiv 1pmod p.$$
Hence $text{ord}_{p}(n)mid 3$. Since $nequiv 2notequiv 1pmod p$ we know that $text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $text{ord}_m(x)mid phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 21:07









Will FisherWill Fisher

4,04811032




4,04811032












  • $begingroup$
    So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:23






  • 2




    $begingroup$
    @MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:24












  • $begingroup$
    It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:51






  • 2




    $begingroup$
    @MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:53










  • $begingroup$
    Great!! Thanks a lot!! :-)
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:59


















  • $begingroup$
    So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:23






  • 2




    $begingroup$
    @MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:24












  • $begingroup$
    It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:51






  • 2




    $begingroup$
    @MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
    $endgroup$
    – Will Fisher
    Dec 15 '18 at 21:53










  • $begingroup$
    Great!! Thanks a lot!! :-)
    $endgroup$
    – Mary Star
    Dec 15 '18 at 21:59
















$begingroup$
So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
$endgroup$
– Mary Star
Dec 15 '18 at 21:23




$begingroup$
So, in this case we have that $text{ord}_p(n)mid phi(p)Rightarrow 3mid p-1 Rightarrow p-1=3k Rightarrow p=3k+1$. That means that $pequiv 1pmod 3$. Any odd divisor will be a product of odd prime divisors, right? So we have to show that the product of terms in the form $3m+1$ will be again in that form, right?
$endgroup$
– Mary Star
Dec 15 '18 at 21:23




2




2




$begingroup$
@MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
$endgroup$
– Will Fisher
Dec 15 '18 at 21:24






$begingroup$
@MaryStar Exactly, but what is a product of things $equiv 1pmod 3$ congruent to mod 3?
$endgroup$
– Will Fisher
Dec 15 '18 at 21:24














$begingroup$
It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
$endgroup$
– Mary Star
Dec 15 '18 at 21:51




$begingroup$
It is congruent to $1$ modulo $3$, since let $aequiv bequiv 1bmod 3$ then $abbmod 3equiv (abmod 3 )(bbmod 3)bmod 3equiv 1bmod 3$, right?
$endgroup$
– Mary Star
Dec 15 '18 at 21:51




2




2




$begingroup$
@MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
$endgroup$
– Will Fisher
Dec 15 '18 at 21:53




$begingroup$
@MaryStar Yep. And like you said, odd divisors are products of odd primes, all of which we know are congruent to 1 mod 3.
$endgroup$
– Will Fisher
Dec 15 '18 at 21:53












$begingroup$
Great!! Thanks a lot!! :-)
$endgroup$
– Mary Star
Dec 15 '18 at 21:59




$begingroup$
Great!! Thanks a lot!! :-)
$endgroup$
– Mary Star
Dec 15 '18 at 21:59


















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