Large Prime Factors!

The expression can be rewrite as

$1^{21}+2^{21}+4^{21}$

We use Vieta's Formula to construct a new equation with roots $p,q,r$ where $p=1^3, q=2^3, r=4^3$ , we will have

$x^3+73x^2+584x+512=0$

Similarly, we need to find the smallest prime factor of

$P_7=p^7+q^7+r^7$

By Newton's Sum ,

$P_1-73=0$

$P_2-73P_1+1168=0$

$P_3-73P_2+584P_1-1536=0$

$P_4-73P_3+584P_2-512P_1=0$

$P_5-73P_4+584P_3-512P_2=0$

$P_6-73P_5+584P_4-512P_3=0$

$P_7-73P_6+584P_5-512P_4=0$

From the last equation, we know that $-73P_6$ and $584P_5$ are both divisible by $73$ . The only left to investigate is $512P_4$ .

From the forth and first equation, which states that $P_1$ is divisible by $73$ , we know that $P_4$ is divisible by $73$ .

Conclude everything above, we get that $P_7$ is divisible by $73$ .

the expression can be expressed as $2^{63}$ -1 / $2^{21}$ -1.
$2^{9}$ -1 is a factor of $2^{63}$ -1 but not $2^{21}$ -1 ( $2^{n}$ -1 is a factor of $2^{an}$ -1).
$2^{9}$ -1= 511 = 7(73).
7= $2^{3}$ -1 is a factor of $2^{21}$ -1 however, hence 73 is not a factor.
hence after division 73 is a factor of the expression.

This is not an infallible or efficient method for finding such prime factors, but this one worked for me. Let ${ a=2 }^{ 3 }$ so that we have $1+{ a }^{ 7 }+{ a }^{ 14 }$ . We note that $1+{ a }+{ a }^{ 2 }=73$ , a prime number, so we hope that $1+{ a }+{ a }^{ 2 }$ divides into $1+{ a }^{ 7 }+{ a }^{ 14 }$ . After some foodling around, we see that it does, i..e $(1+a+{ a }^{ 2 })((1+{ a }^{ 3 }+{ a }^{ 6 }+{ a }^{ 9 }+{ a }^{ 12 })-(a+{ a }^{ 4 }+{ a }^{ 8 }+{ a }^{ 11 }))=(1+{ a }^{ 7 }+{ a }^{ 14 })$ .
And so, I entered "73" as the solution to see what happens.

It's one thing to take a stab at a solution, quite another to prove beyond a doubt that it is indeed the correct one.

The number is $\dfrac{2^{63}-1}{2^{21}-1}$ . Since $x^n - 1 = \prod_{d|n} \Phi_d(x)$ where $\Phi_d$ is the $d$ th cyclotomic polynomial, we get that the number is $\Phi_9(2)\Phi_{63}(2)$ .

Now $\Phi_9(x) = x^6+x^3+1$ , so $\Phi_9(2) = \fbox{73}$ . The stipulations of the problem allow us to stop there and be done.

the following expression 1 + 2^21 + 4^21 can be restated as:

1^3 + (2^3)^7 + (4^3)^7 which is divisible by (1^3 + 2^3 + 4^3 = 73)

thus, the answer is 73

Since $n=1+2^{21}+4^{21}$ is a number of the form $x^2+x+1$ , $-3$ is a quadratic residue for every prime divisor of $n$ , hence all the prime divisors of $n$ are numbers of the form $3k+1$ , and we have only to check divisibility by the elements of the following set: $\{13,19,31,37,43,61,67,73,79,97\}$ . We notice that $p\mid n$ implies $p\mid x^3-1 = 2^{63}-1$ , or $2^{63}\equiv 1\pmod{p}$ . We rule out $13$ since $2^{63}\equiv 2^3\not\equiv 1\pmod{13}$ ; we rule out $19$ since $2^{63}-1\equiv 2^9-1\not\equiv 1\pmod{19}$ and so on, till $73$ . The order of $2$ mod $73$ is just $9$ , so $n\equiv 1+2^3+2^6\equiv 0\pmod{73}$ .

Let $\Phi_n(x)$ denote the $n$ th cyclotomic polynomial.

Then number in question is $\Phi_3\left(2^{21}\right)=\Phi_9\left(2^7\right)=\Phi_9(2)\Phi_{63}(2).$

Thus, $\Phi_9(2)=2^6+2^3+1=73$ is a prime divisor. Since it has two digits, we're done.

I used Mersenne numbers as my guide. Others below used the same idea if not the name. The number given is equivalent to M63 / M21. M63 is divisible by both M7 and M9. M21 is divisible by both M7 and M3. So the number M9 / M3 should divide evenly into M63 / M21. M9 / M3 = 2^6 + 2 ^3 +1 = 73. So 73 is a factor of the original number given.

I know this isn't a solution or a hint, but what I did was factorized (difference of 2 cubes), then got one expression over another. That should get you somewhere...