Is $frac{2^{3510times2}-1}{218^2-1}$ prime? [closed]
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Is $dfrac{2^{3510times2}-1}{218^2-1}$ prime? That it is not prime, can be it proven only by the brute force of a computer or can be proven theoretically? And which are the factors? And what can be said about the factors?
number-theory
closed as off-topic by Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos Nov 25 at 12:21
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Is $dfrac{2^{3510times2}-1}{218^2-1}$ prime? That it is not prime, can be it proven only by the brute force of a computer or can be proven theoretically? And which are the factors? And what can be said about the factors?
number-theory
closed as off-topic by Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos Nov 25 at 12:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10
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up vote
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down vote
favorite
Is $dfrac{2^{3510times2}-1}{218^2-1}$ prime? That it is not prime, can be it proven only by the brute force of a computer or can be proven theoretically? And which are the factors? And what can be said about the factors?
number-theory
Is $dfrac{2^{3510times2}-1}{218^2-1}$ prime? That it is not prime, can be it proven only by the brute force of a computer or can be proven theoretically? And which are the factors? And what can be said about the factors?
number-theory
number-theory
edited Nov 23 at 17:34
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 23 at 17:03
user613967
closed as off-topic by Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos Nov 25 at 12:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos Nov 25 at 12:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Lord Shark the Unknown, Did, Brahadeesh, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10
add a comment |
2
Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10
2
2
Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10
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3 Answers
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up vote
2
down vote
Facts:
- $6 mid kimplies 9mid(2^k-1)$
- $6 mid (2times3510)$
- $9 mid (2^{2times3510}-1)$
- $9 notmid (218^2-1)$
- $3 mid frac{2^{2times3510}-1}{218^2-1}$
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1
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$frac{2^{(3510*2)}-1}{218^{2}-1} = frac{(2^{2})^{3510}-1}{(218+1)(218-1)} = frac{(4)^{3510}-1}{(218+1)(218-1)} $
Lets look on:
$(4)^{3510}-1= (4)^{3510}-1^{3510}$
From factoring rules:
$a^n – b^n = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
$(4-1)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})=$
$ 3(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
This number will always be divisible by 3.
In the denominator, you have a multiplication of two odd numbers. And one of them is divisible by 3 (219).
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
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The factors of $218^2 - 1=(218 +1)(218-1) = 219*217=3*73*7*31$ are easy to find.
Presumably These are also factors of $2^{3510*2}-1=2^{13*3^3*5*4}-1$ (what a weird way of putting it: why $3510*2$?) of which any $2^k -1; k|7020$ is a factor.
But what factors does $2^{13*3^3*5*4}-1$ have which aren't factors of $218^2 - 1= 219*217=3*73*7*31$
Well, $2^k -1; k|7020$ is a facto so $2^4 - 1 = 15 =3*5$ is a factor. The $3$ "divides out" because it is also a factor of $218^2 - 1$ but that $5$ does not. So $5|2^{13*3^3*5*4}-1$ but $5notmid 218^2 - 1$. So your number is not a prime. (And we still haven't verified it is an integer.
To be slightly but not completely more exhaustive some of our factors are
$2^4 - 1 = (2^2-1)(2^2+1) = 3, 5$
As $13*3^3*5=M$ is odd we also have $2^{13*3^3*5*4}-1 = (2^4+1)(2^{4(M-1)}-2^{4(M-2)} + .... + 2^4 - 1)$.
So $2^4 +1 = 17$ is another factor.
$2^{3^3 -1}= (2^9 -1)(2^{18}+2^9 + 1)=(2^3-1)(2^6 + 2^3 +1)(2^{18} + 2^9 + 1)$ so $2^3 - 1= 7$ is a factor that "divides out. $2^6 + 2^3 + 1 = 64+8+1=73$ is a factor that divides out. And thank goodness for calculators we can see $2^{18} + 2^9 + 1 = 262657$ is a factor. ANd thank goodness for google I can see that $262657$ is prime.
$2^5 -1 = 31$ is our final factor that divides out so our number is an integer... which has $5$ and $17$ as factors.
$2^{13} - 1 = 8191$ (which is apparently prime) is a factor.
We haven't fully factored (HUGE understatement) but we have found $5*17*8191*262657$ divide your number. (According to my calculator we still have $1.956684844777819022625663883156e+2097$ as a factor and... screw that. It's not prime and we have $4$ of it's factors...That's enough of an answer.)
====
Hmm, A simple thing I didn't take into account was that combining prime factors. As $6|13*3^3*5*4$ whe have $2^6 - 1=63=7*9$ divides our numerator so $frac 93=3$ divides our value.
Other factors I did not consider were $2^{20} - 1=(2^{10}+1)(2^{5}-1)(2^5 + 1)$ so I hadn't consider that $2^5 + 1=3*11$ and $11$ is a factor. $2^{10} +1 = 1025 = 41*25$ so $25$ and $41$ are factors.
$2^{4*3^3}-1$ will have factors of $2^3 pm 1; 2^9pm 1;2^{27}pm 1$ etc which give $511$ as a factor.
$2^{5*3}-1$ gives $2^{15} -1$ gives $7*31*151$ as factors and so on.
Clearly the numerator has many more (even accounting for duplicates) factors than the denominator does.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Facts:
- $6 mid kimplies 9mid(2^k-1)$
- $6 mid (2times3510)$
- $9 mid (2^{2times3510}-1)$
- $9 notmid (218^2-1)$
- $3 mid frac{2^{2times3510}-1}{218^2-1}$
add a comment |
up vote
2
down vote
Facts:
- $6 mid kimplies 9mid(2^k-1)$
- $6 mid (2times3510)$
- $9 mid (2^{2times3510}-1)$
- $9 notmid (218^2-1)$
- $3 mid frac{2^{2times3510}-1}{218^2-1}$
add a comment |
up vote
2
down vote
up vote
2
down vote
Facts:
- $6 mid kimplies 9mid(2^k-1)$
- $6 mid (2times3510)$
- $9 mid (2^{2times3510}-1)$
- $9 notmid (218^2-1)$
- $3 mid frac{2^{2times3510}-1}{218^2-1}$
Facts:
- $6 mid kimplies 9mid(2^k-1)$
- $6 mid (2times3510)$
- $9 mid (2^{2times3510}-1)$
- $9 notmid (218^2-1)$
- $3 mid frac{2^{2times3510}-1}{218^2-1}$
answered Nov 23 at 17:31
Federico
4,192512
4,192512
add a comment |
add a comment |
up vote
1
down vote
$frac{2^{(3510*2)}-1}{218^{2}-1} = frac{(2^{2})^{3510}-1}{(218+1)(218-1)} = frac{(4)^{3510}-1}{(218+1)(218-1)} $
Lets look on:
$(4)^{3510}-1= (4)^{3510}-1^{3510}$
From factoring rules:
$a^n – b^n = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
$(4-1)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})=$
$ 3(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
This number will always be divisible by 3.
In the denominator, you have a multiplication of two odd numbers. And one of them is divisible by 3 (219).
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
add a comment |
up vote
1
down vote
$frac{2^{(3510*2)}-1}{218^{2}-1} = frac{(2^{2})^{3510}-1}{(218+1)(218-1)} = frac{(4)^{3510}-1}{(218+1)(218-1)} $
Lets look on:
$(4)^{3510}-1= (4)^{3510}-1^{3510}$
From factoring rules:
$a^n – b^n = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
$(4-1)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})=$
$ 3(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
This number will always be divisible by 3.
In the denominator, you have a multiplication of two odd numbers. And one of them is divisible by 3 (219).
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
add a comment |
up vote
1
down vote
up vote
1
down vote
$frac{2^{(3510*2)}-1}{218^{2}-1} = frac{(2^{2})^{3510}-1}{(218+1)(218-1)} = frac{(4)^{3510}-1}{(218+1)(218-1)} $
Lets look on:
$(4)^{3510}-1= (4)^{3510}-1^{3510}$
From factoring rules:
$a^n – b^n = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
$(4-1)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})=$
$ 3(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
This number will always be divisible by 3.
In the denominator, you have a multiplication of two odd numbers. And one of them is divisible by 3 (219).
$frac{2^{(3510*2)}-1}{218^{2}-1} = frac{(2^{2})^{3510}-1}{(218+1)(218-1)} = frac{(4)^{3510}-1}{(218+1)(218-1)} $
Lets look on:
$(4)^{3510}-1= (4)^{3510}-1^{3510}$
From factoring rules:
$a^n – b^n = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
$(4-1)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})=$
$ 3(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + ··· + ab^{n – 2} + b^{n – 1})$
This number will always be divisible by 3.
In the denominator, you have a multiplication of two odd numbers. And one of them is divisible by 3 (219).
answered Nov 23 at 18:02
violettagold
216
216
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
add a comment |
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
But if the numerator and denominator divide by $3$ that doesn't mean the resulting value has $3$ as a factor unless you can show the numerator has a higher power of $3$ as a divisor.
– fleablood
Nov 23 at 18:16
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
Yes, you are right. But if we take (a^(n-1) + b^(n-1)) which are inside parentheses, we can do all the process again and then to get 9 in the numerator.
– violettagold
Nov 23 at 21:55
add a comment |
up vote
0
down vote
The factors of $218^2 - 1=(218 +1)(218-1) = 219*217=3*73*7*31$ are easy to find.
Presumably These are also factors of $2^{3510*2}-1=2^{13*3^3*5*4}-1$ (what a weird way of putting it: why $3510*2$?) of which any $2^k -1; k|7020$ is a factor.
But what factors does $2^{13*3^3*5*4}-1$ have which aren't factors of $218^2 - 1= 219*217=3*73*7*31$
Well, $2^k -1; k|7020$ is a facto so $2^4 - 1 = 15 =3*5$ is a factor. The $3$ "divides out" because it is also a factor of $218^2 - 1$ but that $5$ does not. So $5|2^{13*3^3*5*4}-1$ but $5notmid 218^2 - 1$. So your number is not a prime. (And we still haven't verified it is an integer.
To be slightly but not completely more exhaustive some of our factors are
$2^4 - 1 = (2^2-1)(2^2+1) = 3, 5$
As $13*3^3*5=M$ is odd we also have $2^{13*3^3*5*4}-1 = (2^4+1)(2^{4(M-1)}-2^{4(M-2)} + .... + 2^4 - 1)$.
So $2^4 +1 = 17$ is another factor.
$2^{3^3 -1}= (2^9 -1)(2^{18}+2^9 + 1)=(2^3-1)(2^6 + 2^3 +1)(2^{18} + 2^9 + 1)$ so $2^3 - 1= 7$ is a factor that "divides out. $2^6 + 2^3 + 1 = 64+8+1=73$ is a factor that divides out. And thank goodness for calculators we can see $2^{18} + 2^9 + 1 = 262657$ is a factor. ANd thank goodness for google I can see that $262657$ is prime.
$2^5 -1 = 31$ is our final factor that divides out so our number is an integer... which has $5$ and $17$ as factors.
$2^{13} - 1 = 8191$ (which is apparently prime) is a factor.
We haven't fully factored (HUGE understatement) but we have found $5*17*8191*262657$ divide your number. (According to my calculator we still have $1.956684844777819022625663883156e+2097$ as a factor and... screw that. It's not prime and we have $4$ of it's factors...That's enough of an answer.)
====
Hmm, A simple thing I didn't take into account was that combining prime factors. As $6|13*3^3*5*4$ whe have $2^6 - 1=63=7*9$ divides our numerator so $frac 93=3$ divides our value.
Other factors I did not consider were $2^{20} - 1=(2^{10}+1)(2^{5}-1)(2^5 + 1)$ so I hadn't consider that $2^5 + 1=3*11$ and $11$ is a factor. $2^{10} +1 = 1025 = 41*25$ so $25$ and $41$ are factors.
$2^{4*3^3}-1$ will have factors of $2^3 pm 1; 2^9pm 1;2^{27}pm 1$ etc which give $511$ as a factor.
$2^{5*3}-1$ gives $2^{15} -1$ gives $7*31*151$ as factors and so on.
Clearly the numerator has many more (even accounting for duplicates) factors than the denominator does.
add a comment |
up vote
0
down vote
The factors of $218^2 - 1=(218 +1)(218-1) = 219*217=3*73*7*31$ are easy to find.
Presumably These are also factors of $2^{3510*2}-1=2^{13*3^3*5*4}-1$ (what a weird way of putting it: why $3510*2$?) of which any $2^k -1; k|7020$ is a factor.
But what factors does $2^{13*3^3*5*4}-1$ have which aren't factors of $218^2 - 1= 219*217=3*73*7*31$
Well, $2^k -1; k|7020$ is a facto so $2^4 - 1 = 15 =3*5$ is a factor. The $3$ "divides out" because it is also a factor of $218^2 - 1$ but that $5$ does not. So $5|2^{13*3^3*5*4}-1$ but $5notmid 218^2 - 1$. So your number is not a prime. (And we still haven't verified it is an integer.
To be slightly but not completely more exhaustive some of our factors are
$2^4 - 1 = (2^2-1)(2^2+1) = 3, 5$
As $13*3^3*5=M$ is odd we also have $2^{13*3^3*5*4}-1 = (2^4+1)(2^{4(M-1)}-2^{4(M-2)} + .... + 2^4 - 1)$.
So $2^4 +1 = 17$ is another factor.
$2^{3^3 -1}= (2^9 -1)(2^{18}+2^9 + 1)=(2^3-1)(2^6 + 2^3 +1)(2^{18} + 2^9 + 1)$ so $2^3 - 1= 7$ is a factor that "divides out. $2^6 + 2^3 + 1 = 64+8+1=73$ is a factor that divides out. And thank goodness for calculators we can see $2^{18} + 2^9 + 1 = 262657$ is a factor. ANd thank goodness for google I can see that $262657$ is prime.
$2^5 -1 = 31$ is our final factor that divides out so our number is an integer... which has $5$ and $17$ as factors.
$2^{13} - 1 = 8191$ (which is apparently prime) is a factor.
We haven't fully factored (HUGE understatement) but we have found $5*17*8191*262657$ divide your number. (According to my calculator we still have $1.956684844777819022625663883156e+2097$ as a factor and... screw that. It's not prime and we have $4$ of it's factors...That's enough of an answer.)
====
Hmm, A simple thing I didn't take into account was that combining prime factors. As $6|13*3^3*5*4$ whe have $2^6 - 1=63=7*9$ divides our numerator so $frac 93=3$ divides our value.
Other factors I did not consider were $2^{20} - 1=(2^{10}+1)(2^{5}-1)(2^5 + 1)$ so I hadn't consider that $2^5 + 1=3*11$ and $11$ is a factor. $2^{10} +1 = 1025 = 41*25$ so $25$ and $41$ are factors.
$2^{4*3^3}-1$ will have factors of $2^3 pm 1; 2^9pm 1;2^{27}pm 1$ etc which give $511$ as a factor.
$2^{5*3}-1$ gives $2^{15} -1$ gives $7*31*151$ as factors and so on.
Clearly the numerator has many more (even accounting for duplicates) factors than the denominator does.
add a comment |
up vote
0
down vote
up vote
0
down vote
The factors of $218^2 - 1=(218 +1)(218-1) = 219*217=3*73*7*31$ are easy to find.
Presumably These are also factors of $2^{3510*2}-1=2^{13*3^3*5*4}-1$ (what a weird way of putting it: why $3510*2$?) of which any $2^k -1; k|7020$ is a factor.
But what factors does $2^{13*3^3*5*4}-1$ have which aren't factors of $218^2 - 1= 219*217=3*73*7*31$
Well, $2^k -1; k|7020$ is a facto so $2^4 - 1 = 15 =3*5$ is a factor. The $3$ "divides out" because it is also a factor of $218^2 - 1$ but that $5$ does not. So $5|2^{13*3^3*5*4}-1$ but $5notmid 218^2 - 1$. So your number is not a prime. (And we still haven't verified it is an integer.
To be slightly but not completely more exhaustive some of our factors are
$2^4 - 1 = (2^2-1)(2^2+1) = 3, 5$
As $13*3^3*5=M$ is odd we also have $2^{13*3^3*5*4}-1 = (2^4+1)(2^{4(M-1)}-2^{4(M-2)} + .... + 2^4 - 1)$.
So $2^4 +1 = 17$ is another factor.
$2^{3^3 -1}= (2^9 -1)(2^{18}+2^9 + 1)=(2^3-1)(2^6 + 2^3 +1)(2^{18} + 2^9 + 1)$ so $2^3 - 1= 7$ is a factor that "divides out. $2^6 + 2^3 + 1 = 64+8+1=73$ is a factor that divides out. And thank goodness for calculators we can see $2^{18} + 2^9 + 1 = 262657$ is a factor. ANd thank goodness for google I can see that $262657$ is prime.
$2^5 -1 = 31$ is our final factor that divides out so our number is an integer... which has $5$ and $17$ as factors.
$2^{13} - 1 = 8191$ (which is apparently prime) is a factor.
We haven't fully factored (HUGE understatement) but we have found $5*17*8191*262657$ divide your number. (According to my calculator we still have $1.956684844777819022625663883156e+2097$ as a factor and... screw that. It's not prime and we have $4$ of it's factors...That's enough of an answer.)
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Hmm, A simple thing I didn't take into account was that combining prime factors. As $6|13*3^3*5*4$ whe have $2^6 - 1=63=7*9$ divides our numerator so $frac 93=3$ divides our value.
Other factors I did not consider were $2^{20} - 1=(2^{10}+1)(2^{5}-1)(2^5 + 1)$ so I hadn't consider that $2^5 + 1=3*11$ and $11$ is a factor. $2^{10} +1 = 1025 = 41*25$ so $25$ and $41$ are factors.
$2^{4*3^3}-1$ will have factors of $2^3 pm 1; 2^9pm 1;2^{27}pm 1$ etc which give $511$ as a factor.
$2^{5*3}-1$ gives $2^{15} -1$ gives $7*31*151$ as factors and so on.
Clearly the numerator has many more (even accounting for duplicates) factors than the denominator does.
The factors of $218^2 - 1=(218 +1)(218-1) = 219*217=3*73*7*31$ are easy to find.
Presumably These are also factors of $2^{3510*2}-1=2^{13*3^3*5*4}-1$ (what a weird way of putting it: why $3510*2$?) of which any $2^k -1; k|7020$ is a factor.
But what factors does $2^{13*3^3*5*4}-1$ have which aren't factors of $218^2 - 1= 219*217=3*73*7*31$
Well, $2^k -1; k|7020$ is a facto so $2^4 - 1 = 15 =3*5$ is a factor. The $3$ "divides out" because it is also a factor of $218^2 - 1$ but that $5$ does not. So $5|2^{13*3^3*5*4}-1$ but $5notmid 218^2 - 1$. So your number is not a prime. (And we still haven't verified it is an integer.
To be slightly but not completely more exhaustive some of our factors are
$2^4 - 1 = (2^2-1)(2^2+1) = 3, 5$
As $13*3^3*5=M$ is odd we also have $2^{13*3^3*5*4}-1 = (2^4+1)(2^{4(M-1)}-2^{4(M-2)} + .... + 2^4 - 1)$.
So $2^4 +1 = 17$ is another factor.
$2^{3^3 -1}= (2^9 -1)(2^{18}+2^9 + 1)=(2^3-1)(2^6 + 2^3 +1)(2^{18} + 2^9 + 1)$ so $2^3 - 1= 7$ is a factor that "divides out. $2^6 + 2^3 + 1 = 64+8+1=73$ is a factor that divides out. And thank goodness for calculators we can see $2^{18} + 2^9 + 1 = 262657$ is a factor. ANd thank goodness for google I can see that $262657$ is prime.
$2^5 -1 = 31$ is our final factor that divides out so our number is an integer... which has $5$ and $17$ as factors.
$2^{13} - 1 = 8191$ (which is apparently prime) is a factor.
We haven't fully factored (HUGE understatement) but we have found $5*17*8191*262657$ divide your number. (According to my calculator we still have $1.956684844777819022625663883156e+2097$ as a factor and... screw that. It's not prime and we have $4$ of it's factors...That's enough of an answer.)
====
Hmm, A simple thing I didn't take into account was that combining prime factors. As $6|13*3^3*5*4$ whe have $2^6 - 1=63=7*9$ divides our numerator so $frac 93=3$ divides our value.
Other factors I did not consider were $2^{20} - 1=(2^{10}+1)(2^{5}-1)(2^5 + 1)$ so I hadn't consider that $2^5 + 1=3*11$ and $11$ is a factor. $2^{10} +1 = 1025 = 41*25$ so $25$ and $41$ are factors.
$2^{4*3^3}-1$ will have factors of $2^3 pm 1; 2^9pm 1;2^{27}pm 1$ etc which give $511$ as a factor.
$2^{5*3}-1$ gives $2^{15} -1$ gives $7*31*151$ as factors and so on.
Clearly the numerator has many more (even accounting for duplicates) factors than the denominator does.
edited Nov 23 at 20:04
answered Nov 23 at 18:07
fleablood
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Hint: if $dmid n$, then $2^d-1mid 2^n-1$. You can't quite use this directly, but it will be of help.
– Wojowu
Nov 23 at 17:06
Your number is divisible by $3$...
– Federico
Nov 23 at 17:11
@Federico Does it? $3$ divides the numerator and the denominator but does a higher power divide that numerator than divides the denominator? (I don't think so, I think that both are divisible by $3^1$, but I could be wrong). But $5$ and $17$ divide the numerator but not the denominator so it can't be prime. (It might not be an integer.)
– fleablood
Nov 23 at 18:10