Is $\frac{2^{3510\times2}-1}{218^2-1}$ prime?

Question

Is $\dfrac{2^{3510\times2}-1}{218^2-1}$ 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?

Answer 1

Facts:

• $6 \mid k\implies 9\mid(2^k-1)$
• $6 \mid (2\times3510)$
• $9 \mid (2^{2\times3510}-1)$
• $9 \not\mid (218^2-1)$
• $3 \mid \frac{2^{2\times3510}-1}{218^2-1}$

Answer 2

$\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).

Answer 3

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 $5\not\mid 218^2 - 1$. 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 $2^3 \pm 1; 2^9\pm 1;2^{27}\pm 1$ 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.