Sum of all the factors of $33333333$ What is the sum of factors of factors of $33333333$ (that's $3$ eight times)?
Here's how I tried to attempt this question yet failed:
Factors of $33333333=3\times11111111=3\times11\times1010101=3\times11\times101\times10001$.
And now I was lost (just before the last step) as I couldn't confirm if 10001 is prime or not.
A solution I got: $10001=11025-1024=105^2-32^2=137\times73$ which are primes. But no way could I have myself figured out that 10001 is difference of those 2 squares.
Please help me in finding an easy way to check whether 10001 is prime or not and what it factors would as they lie too far to check manually, one-by-one, till $\sqrt{10001}=100$ (approx).
Edit: "just before the last step" above refers to the following step I would have done had I known the prime factorization:
Sum of all the factors of  $33333333=(\frac{3^2-1}{3-1})\times(\frac{11^2-1}{11-1})\times(\frac{101^2-1}{101-1})\times(\frac{137^2-1}{137-1})\times(\frac{73^2-1}{73-1})=(3+1)(11+1)(101+1)(137+1)(73+1)$
 A: Fermat's factoring technique is extremely fast for 10001.
$$101^2 - 10001 = 10201 - 10001 = 200$$
$$101\cdot 2 + 1 = 203$$
$$\begin{array} {r|c|l}
200 & 101 & \text{Obv not a square} \\
+203 \\
403 & 102 & \text{Square can't end in 3} \\
+205 \\
608 & 103 & \text{Square can't have digital root 5} \\
+207 \\
815 & 104 & \text{Square can't have digital root 5} \\
+ 209 \\
1024 & 105 & \text{Found It} \\
\end{array}$$

Each odd row $(a,b)$ is $b^2 - n = a$, for example, $104^2 - 10001 = 815$.  Eventually we find $105^2 - 10001 = 1024 = 32^2$, which implies $10001 = 105^2 - 32^2 = (105 + 32)(105 - 32) = 137 \times 73$.
Finding the $a$ is easy, just add.  Ruling out that it is a square requires tricks.  Like a square can only end in 1, 4, 5, 6, 9 (or 00) and only have a digital root of 1, 4, 7 or 9, which rules out 75% of them.  You could make a large table that shows all possible 2 digit endings of squares which would rule out half of the remaining possibilities.  And sometimes, at worst, you'd just have to check if a number is a square by squaring an estimate of it's square root.
A: Honestly, there isn't a very good way to factor 10001 without just brute forcing through every prime up to 100. As you said, the difference of squares is very hard to find, and that method shouldn't be used unless it is extremely obvious (for example, the number 9991). Once you find the primes, you can easily find the sum of all the factors using the method you described in your edit.
A: There's no known fast general algorithm for factoring integers.  (Indeed, the RSA cryptosystem relies on the assumption that factoring is hard.)
One thing that might help somewhat is to write your numbers in base six, which makes it easier to identify primes because there are only two digits that a prime number (other than 2 or 3) can end with: 1 or 5.
$10001_{10} = 114145_6$.  This ends in a 5, so it might be a prime number.  To narrow down possible factors, consult the multiplication table:
1 2 3 4 5
2 4 0 2 4
3 0 3 0 3
4 2 0 4 2
5 4 3 2 1

In order to get a 5 as the last base-6 digit, one factor must end in 1, and the other must end in 5.
The 25 prime numbers less than $\sqrt{10001}$ break down as follows:

*

*11 whose last base-6 digit is 1: 11 (7), 21 (13), 31 (19), 51 (31), 101 (37), 111 (43), 141 (61), 151 (67), 201 (73), 211 (79), 241 (97)

*12 whose last base-6 digit is 5: 5, 15 (11), 25 (17), 35 (23), 45 (29), 105 (41), 115 (47), 125 (53), 135 (59), 155 (71), 215 (83), 225 (89)

*2 special cases that are obviously not factors here: 2, 3

So, instead of trying all 25 primes, you just need to try the 11 "1" primes, if there's a factor, division will get you the corresponding "5" prime.
$$114145_6 = 201_6 \times 345_6$$
$$10001 = 73 \times 137$$
