Proving $10240…02401$ composite











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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.










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  • 3




    What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
    – Surb
    Jul 27 at 9:23






  • 1




    It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
    – MalayTheDynamo
    Jul 27 at 9:24








  • 2




    Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
    – Mark Bennet
    Jul 27 at 9:28






  • 2




    @Servaes The strong Fermat test to base $2$ says it's composite.
    – Daniel Fischer
    Jul 27 at 11:24






  • 4




    @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.)
    – Daniel Fischer
    Jul 27 at 11:50

















up vote
22
down vote

favorite
14












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.










share|cite|improve this question




















  • 3




    What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
    – Surb
    Jul 27 at 9:23






  • 1




    It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
    – MalayTheDynamo
    Jul 27 at 9:24








  • 2




    Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
    – Mark Bennet
    Jul 27 at 9:28






  • 2




    @Servaes The strong Fermat test to base $2$ says it's composite.
    – Daniel Fischer
    Jul 27 at 11:24






  • 4




    @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.)
    – Daniel Fischer
    Jul 27 at 11:50















up vote
22
down vote

favorite
14









up vote
22
down vote

favorite
14






14





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.










share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 9:35

























asked Jul 27 at 9:14









MalayTheDynamo

1,617933




1,617933








  • 3




    What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
    – Surb
    Jul 27 at 9:23






  • 1




    It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
    – MalayTheDynamo
    Jul 27 at 9:24








  • 2




    Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
    – Mark Bennet
    Jul 27 at 9:28






  • 2




    @Servaes The strong Fermat test to base $2$ says it's composite.
    – Daniel Fischer
    Jul 27 at 11:24






  • 4




    @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.)
    – Daniel Fischer
    Jul 27 at 11:50
















  • 3




    What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
    – Surb
    Jul 27 at 9:23






  • 1




    It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
    – MalayTheDynamo
    Jul 27 at 9:24








  • 2




    Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
    – Mark Bennet
    Jul 27 at 9:28






  • 2




    @Servaes The strong Fermat test to base $2$ says it's composite.
    – Daniel Fischer
    Jul 27 at 11:24






  • 4




    @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.)
    – Daniel Fischer
    Jul 27 at 11:50










3




3




What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
– Surb
Jul 27 at 9:23




What does "102400...(2014 times)...002401" mean? It's not clear to me what should be the $...$.
– Surb
Jul 27 at 9:23




1




1




It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
– MalayTheDynamo
Jul 27 at 9:24






It means that $1024$ and $2401$ have $2014text{ '0'}s$ between them.
– MalayTheDynamo
Jul 27 at 9:24






2




2




Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
– Mark Bennet
Jul 27 at 9:28




Am I right in thinking that your number has $2022$ digits of which $2016$ are zero?
– Mark Bennet
Jul 27 at 9:28




2




2




@Servaes The strong Fermat test to base $2$ says it's composite.
– Daniel Fischer
Jul 27 at 11:24




@Servaes The strong Fermat test to base $2$ says it's composite.
– Daniel Fischer
Jul 27 at 11:24




4




4




@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.)
– Daniel Fischer
Jul 27 at 11:50






@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.)
– Daniel Fischer
Jul 27 at 11:50

















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