Actually 3 Problems in 1

First we shall factor $x^{22} - y^{22}$

You get:

$(x-y) (x+y) \left(x^{10}-x^9 y+x^8 y^2-x^7 y^3+x^6 y^4-x^5 y^5+x^4 y^6-x^3 y^7+x^2 y^8-x y^9+y^{10}\right) \left(x^{10}+x^9 y+x^8 y^2+x^7 y^3+x^6 y^4+x^5 y^5+x^4 y^6+x^3 y^7+x^2 y^8+x y^9+y^{10}\right)$

Now, plugging in the values of x and y with 6 and 1, you get $\boxed{5}$ and $\boxed{7}$ for the first two factors.

We need another factor:

Consider $(6^{22}-1)$ . Note that 22 = 23 -1 and 23 is prime

By Fermat's Little theorem,

$6^{22}\equiv6^{23-1}\equiv1 \pmod {23}$

$\\ \implies 23 | (6^{22} - 1)$

So, the other factor is $\boxed{23}$

TL;DR Go to Wolfram|Alpha and type

FactorInteger[6^22-1]

which gives

{{5, 1}, {7, 1}, {23, 1}, {3154757, 1}, {51828151, 1}}

If you liked my solution, please upvote and tell me what is the correct way to write divides in LaTeX

We are looking for the smallest three primes $m$ where $6^{22} \equiv 1\pmod{m}$ .

Obviously, 2 and 3 are not valid options, so we next look to 5.

It so happens that $6 \equiv 6^{22} \equiv 1 \pmod{5} \implies 5 \mid 6^{22}$ , and there we have our first prime.

Logically, the next step would be to look at 7. $6 \equiv -1 \pmod{7} \implies 6^2 \equiv 1\pmod{7} \implies {(6^2)}^{11} \equiv 1 \pmod{7}$ $\implies 7 \mid 6^{22}$ . So we have our second prime.

Note: More generally, if you have a number $a^n-1$ , it would always be divisible by $a-1$ , and if $n$ is an even number, $a+1$ would also be a factor.

If you keep working up the primes (11, 13, etc), you, like me, would realize that you are just wasting your time because it doesn't work out and there is no pattern whatsoever.

So I got bored and turned to Fermat's Little Theorem. $6^{22}=6^{23-1}\equiv 1 \pmod{23} \implies 23\mid 6^{22}$ .

Therefore, the answer is $5+7+23=\boxed{35}$ .

Bash the mods of the first primes and find that the first 3 that work are 5,7,and 23. Sum them to get 35.

Neat trick. I programmed a python prime factorization algorithm to solve.

Checking cases in ascending order gives us 5 and 7 as primefactors. The rest upto 19 can be checked using modular arithematic. for 23, Observe that 22=23-1 i.e. a PRIME! So, Check if fermat's little theorem can be used.

= 131621703842267135 = 5 x 7 x 23 x 163505222164307