Pretty Large

Relevant wiki: Cyclotomic Polynomials

Here's how I solved the problem, making use of $2^a -1 \mid 2^{ab} - 1$ that I was alluding to.

We know that $2^ { 15} - 1 \mid 2^{105 } - 1$ . It is easy to see that $\frac{ 2 ^ { 15} -1 } { ( 2^3 - 1 ) ( 2^ 5 - 1 ) }$ would be a proper factor (not necessarily prime) of the expression. Hence the expression is not prime.

More generally, we know that $(2^{105} - 1 ) = \prod_{d \mid 105 } \Phi_d (2)$ , which gives us 8 factors of 105 (of which $\Phi_1(2) = 1$ ). This expresses $2^ { 105} - 1$ as the product of 7 (not necessarily prime) factors, and we've only eliminated 3 prime factors. The other factors of the expression are $\Phi_{15} (2)$ (which is the term I gave above), $\Phi_{21} (2), \Phi_{35} (2),$ and $\Phi_{105} (2)$ .

---

In particular, this generalizes easily. We didn't need to hunt down a certain factor that worked.

Do you see why I had to use $105 = 3 \times 5 \times 7$ as the product of 3 prime factors? Simply using $15 = 3 \times 5$ would only give us 1 additional factor.

First we show that 49 divides $2^{21}-1$ . Indeed, from $2^{10} = 1024 = 21 \cdot 49 - 5 \equiv -5 \pmod{49}$ we have $2^{21} = 2 \cdot (2^{10})^2 \equiv 2 \cdot (-5)^2 \equiv 50 \equiv 1 \pmod{49}$ , so $2^{21}-1 \equiv 0 \pmod{49}$ . By 21 | 105 we get $2^{21}-1$ | $2^{105}-1$ , hence we can conclude that the power of 7 in the factorisation of $2^{105}-1$ has to be at least 2. Since $2^{105}-1$ divided by 7, 31 and 127 is definitely bigger than 7, furthermore we have proven that it has to be divisible by 7, it follows that it cannot be a prime.

Relevant Wiki : Lifting the Exponent

The number $2^{105}-1$ is divided by $7,31,127$ .

We need to find how many number of primes $7,31,127$ exists in the factorization of $2^{105}-1$

First start with $7$

$2^{105}-1=(2^{15})^7-1$

Herem, $7 |2^{15}-1$

Proof: $2^6 \equiv 1=> 2^{12} \equiv 1 => 2^{15}=2^{12}.2^3 \equiv 1 \pmod{7}$

So, using $LTE$ lemma

$v_p(x^n-y^n)=v_p(x-y) + v_p(n)$

$v_7(2^{105}-1)=v_7(2^{15}-1) + v_7(105)$

$v_7(2^{105}-1) \geq 1+1=2$

So, there is at leat two factors of $7$ are there.

When $2^{105}-1$ is divided by the primes $7,31,127$ there will be at least one $7$ and other quotient part which is achieved by dividing $31,127$ .As $gcd(7,31,127)=1$ So, at least $7$ will be in tact.

Finally the quotient will be looks like $7 \times q$ which can not be prime