Finding the smallest prime factor of $sum_{a=1}^N a^{k}$
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This question is linked to my previous question, but I wanted a clearer explanation.
Suppose we have a huge number of that type with a huge $k$.
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k},$$
and we want to find the smallest prime factor. We want to find the smallest $p$ for which
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k}equiv 0 pmod p.$$
Let's review some facts about $a^{k}pmod p$: by Fermat's Little Theorem $a^{k}equiv 1pmod p$ if $(p-1)mid k∧p∤a$. Otherwise $a^{k}≢1pmod p$ and in the particular case when $p|a⟹a^{k} equiv 0pmod p$. If the exponent is a multiple of $p-1$, the powers becomes equal to 1.
Okay, now it's here where I get stuck. How do I continue to find the smallest prime factor? Should I count each $a^{k} pmod p$? I think that I am missing something but I can't nail it.
prime-numbers modular-arithmetic exponentiation prime-factorization
edited Nov 23 at 11:34
Klangen
1,50811232
asked Sep 3 at 10:02
alienflow
787
|
show 3 more comments
up vote
0
down vote
favorite
This question is linked to my previous question, but I wanted a clearer explanation.
Suppose we have a huge number of that type with a huge $k$.
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k},$$
and we want to find the smallest prime factor. We want to find the smallest $p$ for which
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k}equiv 0 pmod p.$$
Let's review some facts about $a^{k}pmod p$: by Fermat's Little Theorem $a^{k}equiv 1pmod p$ if $(p-1)mid k∧p∤a$. Otherwise $a^{k}≢1pmod p$ and in the particular case when $p|a⟹a^{k} equiv 0pmod p$. If the exponent is a multiple of $p-1$, the powers becomes equal to 1.
Okay, now it's here where I get stuck. How do I continue to find the smallest prime factor? Should I count each $a^{k} pmod p$? I think that I am missing something but I can't nail it.
prime-numbers modular-arithmetic exponentiation prime-factorization
edited Nov 23 at 11:34
Klangen
1,50811232
asked Sep 3 at 10:02
alienflow
787
Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is linked to my previous question, but I wanted a clearer explanation.
Suppose we have a huge number of that type with a huge $k$.
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k},$$
and we want to find the smallest prime factor. We want to find the smallest $p$ for which
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k}equiv 0 pmod p.$$
Let's review some facts about $a^{k}pmod p$: by Fermat's Little Theorem $a^{k}equiv 1pmod p$ if $(p-1)mid k∧p∤a$. Otherwise $a^{k}≢1pmod p$ and in the particular case when $p|a⟹a^{k} equiv 0pmod p$. If the exponent is a multiple of $p-1$, the powers becomes equal to 1.
Okay, now it's here where I get stuck. How do I continue to find the smallest prime factor? Should I count each $a^{k} pmod p$? I think that I am missing something but I can't nail it.
prime-numbers modular-arithmetic exponentiation prime-factorization
edited Nov 23 at 11:34
Klangen
1,50811232
asked Sep 3 at 10:02
alienflow
787
This question is linked to my previous question, but I wanted a clearer explanation.
Suppose we have a huge number of that type with a huge $k$.
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k},$$
and we want to find the smallest prime factor. We want to find the smallest $p$ for which
$$sum_{a=1}^N a^{k} =1^{k}+2^{k}+3^{k}+...+N^{k}equiv 0 pmod p.$$
Let's review some facts about $a^{k}pmod p$: by Fermat's Little Theorem $a^{k}equiv 1pmod p$ if $(p-1)mid k∧p∤a$. Otherwise $a^{k}≢1pmod p$ and in the particular case when $p|a⟹a^{k} equiv 0pmod p$. If the exponent is a multiple of $p-1$, the powers becomes equal to 1.
Okay, now it's here where I get stuck. How do I continue to find the smallest prime factor? Should I count each $a^{k} pmod p$? I think that I am missing something but I can't nail it.
prime-numbers modular-arithmetic exponentiation prime-factorization
prime-numbers modular-arithmetic exponentiation prime-factorization
edited Nov 23 at 11:34
Klangen
1,50811232
asked Sep 3 at 10:02
alienflow
787
edited Nov 23 at 11:34
Klangen
1,50811232
asked Sep 3 at 10:02
alienflow
787
edited Nov 23 at 11:34
Klangen
1,50811232
edited Nov 23 at 11:34
Klangen
1,50811232
edited Nov 23 at 11:34
Klangen
1,50811232
1,50811232
asked Sep 3 at 10:02
alienflow
787
asked Sep 3 at 10:02
alienflow
787
asked Sep 3 at 10:02
alienflow
787
787
Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14
|
show 3 more comments
Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14
Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14
|
show 3 more comments
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Possibly misleading use of the polymorphic $|$ to mean both "such as" and "divides" in cases where either interpretation lead to well formed statements.
– marshal craft
Sep 3 at 10:58
Here I use it mainly to mean that it DIVIDES
– alienflow
Sep 3 at 10:59
Also as i do not know where this question comes from, i assume worst case it is dependent on reiman hypothesis? And you have to divide a whole bunch of times.
– marshal craft
Sep 3 at 11:03
I would start by considering cases, either it is prime or not.
– marshal craft
Sep 3 at 11:04
Also maybe related to euler product. Well seems to be it but only for negative $s$ and for finite sums.
– marshal craft
Sep 3 at 11:14