Proving $10240…02401$ composite
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
$endgroup$
|
show 16 more comments
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
$endgroup$
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
$begingroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
$endgroup$
I got this question recently, and have been unable to solve it.
Prove that $1024underbrace{00 ldotsldots 00}_{2014 text{ times}}2401$ is composite.
I have two different ways in mind.
First is $7^4+400(2^2cdot10^{504})^4$, which looks like Sophie Germain, but I'm not sure what to do with the $400$. Another thought is that this is almost a palindrome, with the order of just two digits interchanged. I'm not sure where to go from there, and if it'd provide any results, but it seems interesting nonetheless.
Please help.
elementary-number-theory contest-math arithmetic
elementary-number-theory contest-math arithmetic
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
edited Jul 27 '18 at 9:35
MalayTheDynamo
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
asked Jul 27 '18 at 9:14
MalayTheDynamoMalayTheDynamo
1,614933
1,614933
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
3
3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50
|
show 16 more comments
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3
$begingroup$
What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
$endgroup$
– Surb
Jul 27 '18 at 9:23
1
$begingroup$
It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
$endgroup$
– MalayTheDynamo
Jul 27 '18 at 9:24
2
$begingroup$
Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
$endgroup$
– Mark Bennet
Jul 27 '18 at 9:28
2
$begingroup$
@Servaes The strong Fermat test to base $2$ says it's composite.
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:24
4
$begingroup$
@Servaes Write $n-1 = 2^kcdot m$ with $m$ odd, and then $$a^{n-1} - 1 = (a^m - 1)prod_{kappa = 0}^{k-1}bigl(a^{2^{kappa} m} + 1bigr)$$ where $1 < a < n-1$. Then $n$ is a strong Fermat probable prime if $n$ divides one of the factors of the product, i.e. $a^m equiv 1 pmod{n}$ or there is a $kappa in {0,dotsc, k-1}$ such that $a^{2^{kappa}m} equiv -1pmod{n}$. That's the test Miller-Rabin is composed of. In this case, $a = 2$ shows it's composite. (Actually already the ordinary Fermat test shows that here.)
$endgroup$
– Daniel Fischer♦
Jul 27 '18 at 11:50