Prove $3^x+9^x+1$ is divisible by 13 if $x=3n+1$












0












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I don't know how to start thinking about this problem, I was going to try to prove it by induction, but I think I'm on the wrong path. Any hints?






number-theory







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$endgroup$












  • $begingroup$
    The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:26










  • $begingroup$
    "It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:29












  • $begingroup$
    $3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:33










  • $begingroup$
    Oh, of course, thank you.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:34
















0












$begingroup$


I don't know how to start thinking about this problem, I was going to try to prove it by induction, but I think I'm on the wrong path. Any hints?






number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:26










  • $begingroup$
    "It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:29












  • $begingroup$
    $3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:33










  • $begingroup$
    Oh, of course, thank you.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:34














0












0








0





$begingroup$


I don't know how to start thinking about this problem, I was going to try to prove it by induction, but I think I'm on the wrong path. Any hints?






number-theory







share|cite|improve this question









$endgroup$




I don't know how to start thinking about this problem, I was going to try to prove it by induction, but I think I'm on the wrong path. Any hints?






number-theory




number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 2:14









LowkeyLowkey

728




728












  • $begingroup$
    The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:26










  • $begingroup$
    "It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:29












  • $begingroup$
    $3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:33










  • $begingroup$
    Oh, of course, thank you.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:34


















  • $begingroup$
    The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:26










  • $begingroup$
    "It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:29












  • $begingroup$
    $3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 3:33










  • $begingroup$
    Oh, of course, thank you.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:34
















$begingroup$
The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 3:26




$begingroup$
The order of $3$ in $mathbb{Z}/(13mathbb{Z})^*$ is $3$ since $3^3=27equiv 1pmod{13}$. It follows that the remainder of $1+3^x+9^xpmod{13}$ only depends on $xpmod{3}$ and you just have to test $xin{0,1,2}$ to get that $1+3^x+9^xequiv 0pmod{13}$ iff $xnotequiv 0pmod{3}$.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 3:26












$begingroup$
"It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
$endgroup$
– Lowkey
Dec 7 '18 at 3:29






$begingroup$
"It follows that the remainder only depends on x" could you explain this a little more? I understand that 3 has order 3 in that ring, but I don't get how to use that fact.
$endgroup$
– Lowkey
Dec 7 '18 at 3:29














$begingroup$
$3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 3:33




$begingroup$
$3^{x+3k}equiv 3^{x}pmod{13}$, hence $3^{x}pmod{13}$ is the same thing as $3^{xpmod{3}}pmod{13}$.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 3:33












$begingroup$
Oh, of course, thank you.
$endgroup$
– Lowkey
Dec 7 '18 at 3:34




$begingroup$
Oh, of course, thank you.
$endgroup$
– Lowkey
Dec 7 '18 at 3:34










3 Answers
3






active

oldest

votes


















1












$begingroup$

Note that $3^{3n}equiv 1 bmod 13$, and that $9^{3n}equiv 1 bmod 13$.



Edit: Typo!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:25












  • $begingroup$
    @Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:26












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:30





















1












$begingroup$

Below put $,x = 3,, {A,B,C} = {2n,n,0} $ for any $,nnotequiv 0pmod{!3}$



Lemma $ x^{2}!+!x!+!1mid x^A! +! x^B! +! x^C $ if $ {A,B,C}equiv {2,1,0}pmod{!3}. $



Proof $ $ Special case of a simple proof in this answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many further examples are in this list.
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 3:41










  • $begingroup$
    Thank you! I was actually confused by the other comment, and this was very helpful.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:44



















0












$begingroup$

Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$



If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity



Consider $n=3m+1,3m+2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:24










  • $begingroup$
    @RandomMathGuy, Yes we have derived a generalization. Set $y=3$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:29










  • $begingroup$
    @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:33










  • $begingroup$
    @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:36






  • 1




    $begingroup$
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:00













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that $3^{3n}equiv 1 bmod 13$, and that $9^{3n}equiv 1 bmod 13$.



Edit: Typo!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:25












  • $begingroup$
    @Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:26












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:30


















1












$begingroup$

Note that $3^{3n}equiv 1 bmod 13$, and that $9^{3n}equiv 1 bmod 13$.



Edit: Typo!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:25












  • $begingroup$
    @Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:26












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:30
















1












1








1





$begingroup$

Note that $3^{3n}equiv 1 bmod 13$, and that $9^{3n}equiv 1 bmod 13$.



Edit: Typo!






share|cite|improve this answer









$endgroup$



Note that $3^{3n}equiv 1 bmod 13$, and that $9^{3n}equiv 1 bmod 13$.



Edit: Typo!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 2:17









RandomMathGuyRandomMathGuy

1192




1192












  • $begingroup$
    Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:25












  • $begingroup$
    @Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:26












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:30




















  • $begingroup$
    Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:25












  • $begingroup$
    @Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:26












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Lowkey
    Dec 7 '18 at 2:30


















$begingroup$
Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
$endgroup$
– Lowkey
Dec 7 '18 at 2:25






$begingroup$
Oh, so $3^(3n+1)≡3 mod13$ and $9^(3n+1)≡9 mod13$, so the sum is $3+9+1=0 mod13$?
$endgroup$
– Lowkey
Dec 7 '18 at 2:25














$begingroup$
@Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:26






$begingroup$
@Lowkey Bingo! Anytime you work with modular arithmetic, the key is usually going to be in exploiting some kind of cyclic pattern. This even applies to iterated modular exponentiation.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:26














$begingroup$
Thank you for your help!
$endgroup$
– Lowkey
Dec 7 '18 at 2:30






$begingroup$
Thank you for your help!
$endgroup$
– Lowkey
Dec 7 '18 at 2:30













1












$begingroup$

Below put $,x = 3,, {A,B,C} = {2n,n,0} $ for any $,nnotequiv 0pmod{!3}$



Lemma $ x^{2}!+!x!+!1mid x^A! +! x^B! +! x^C $ if $ {A,B,C}equiv {2,1,0}pmod{!3}. $



Proof $ $ Special case of a simple proof in this answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many further examples are in this list.
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 3:41










  • $begingroup$
    Thank you! I was actually confused by the other comment, and this was very helpful.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:44
















1












$begingroup$

Below put $,x = 3,, {A,B,C} = {2n,n,0} $ for any $,nnotequiv 0pmod{!3}$



Lemma $ x^{2}!+!x!+!1mid x^A! +! x^B! +! x^C $ if $ {A,B,C}equiv {2,1,0}pmod{!3}. $



Proof $ $ Special case of a simple proof in this answer.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Many further examples are in this list.
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 3:41










  • $begingroup$
    Thank you! I was actually confused by the other comment, and this was very helpful.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:44














1












1








1





$begingroup$

Below put $,x = 3,, {A,B,C} = {2n,n,0} $ for any $,nnotequiv 0pmod{!3}$



Lemma $ x^{2}!+!x!+!1mid x^A! +! x^B! +! x^C $ if $ {A,B,C}equiv {2,1,0}pmod{!3}. $



Proof $ $ Special case of a simple proof in this answer.






share|cite|improve this answer











$endgroup$



Below put $,x = 3,, {A,B,C} = {2n,n,0} $ for any $,nnotequiv 0pmod{!3}$



Lemma $ x^{2}!+!x!+!1mid x^A! +! x^B! +! x^C $ if $ {A,B,C}equiv {2,1,0}pmod{!3}. $



Proof $ $ Special case of a simple proof in this answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 3:48

























answered Dec 7 '18 at 3:39









Bill DubuqueBill Dubuque

210k29191639




210k29191639












  • $begingroup$
    Many further examples are in this list.
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 3:41










  • $begingroup$
    Thank you! I was actually confused by the other comment, and this was very helpful.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:44


















  • $begingroup$
    Many further examples are in this list.
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 3:41










  • $begingroup$
    Thank you! I was actually confused by the other comment, and this was very helpful.
    $endgroup$
    – Lowkey
    Dec 7 '18 at 3:44
















$begingroup$
Many further examples are in this list.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 3:41




$begingroup$
Many further examples are in this list.
$endgroup$
– Bill Dubuque
Dec 7 '18 at 3:41












$begingroup$
Thank you! I was actually confused by the other comment, and this was very helpful.
$endgroup$
– Lowkey
Dec 7 '18 at 3:44




$begingroup$
Thank you! I was actually confused by the other comment, and this was very helpful.
$endgroup$
– Lowkey
Dec 7 '18 at 3:44











0












$begingroup$

Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$



If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity



Consider $n=3m+1,3m+2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:24










  • $begingroup$
    @RandomMathGuy, Yes we have derived a generalization. Set $y=3$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:29










  • $begingroup$
    @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:33










  • $begingroup$
    @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:36






  • 1




    $begingroup$
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:00


















0












$begingroup$

Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$



If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity



Consider $n=3m+1,3m+2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:24










  • $begingroup$
    @RandomMathGuy, Yes we have derived a generalization. Set $y=3$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:29










  • $begingroup$
    @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:33










  • $begingroup$
    @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:36






  • 1




    $begingroup$
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:00
















0












0








0





$begingroup$

Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$



If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity



Consider $n=3m+1,3m+2$






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$endgroup$



Proposition $$y^{2n}+y^n+1$$ is divisible by $y^2+y+1$ if integer $n$ is not divisible by $3$



If $y^2+y+1=0,y=w,w^2$ where $w$ is a complex cube root of unity



Consider $n=3m+1,3m+2$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 2:22









lab bhattacharjeelab bhattacharjee

225k15157275




225k15157275












  • $begingroup$
    I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:24










  • $begingroup$
    @RandomMathGuy, Yes we have derived a generalization. Set $y=3$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:29










  • $begingroup$
    @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:33










  • $begingroup$
    @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:36






  • 1




    $begingroup$
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:00




















  • $begingroup$
    I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:24










  • $begingroup$
    @RandomMathGuy, Yes we have derived a generalization. Set $y=3$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:29










  • $begingroup$
    @lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:33










  • $begingroup$
    @RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 2:36






  • 1




    $begingroup$
    I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 3:00


















$begingroup$
I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:24




$begingroup$
I am very confused by what you are trying to say with your answer... that we should prove the stronger statement? That there is a cube root of unity involved in what we are looking at?
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:24












$begingroup$
@RandomMathGuy, Yes we have derived a generalization. Set $y=3$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 2:29




$begingroup$
@RandomMathGuy, Yes we have derived a generalization. Set $y=3$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 2:29












$begingroup$
@lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:33




$begingroup$
@lab_bhattacharjee Except you didn't actually derive it. You just stated the proposition, and then the definition of a cube root of unity.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:33












$begingroup$
@RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 2:36




$begingroup$
@RandomMathGuy, follow the last line: for $n=3m+1$ $$w^{2(3m+1)}+w^{3m+1}+1=0$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 2:36




1




1




$begingroup$
I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:00






$begingroup$
I find it very unfortunate you seem to be unwilling to share this line of reasoning with the community, so I will; it is really clever to recognize that $3$ and $9$ play the role of cube roots of unity $bmod 13$, and that because of this so long as their exponents don't cause them to degenerate we will have that the sum of all three roots will be congruent to $0$, giving us the divisibility property that we want.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:00




















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