Factorize FTW!

The key thing here is noting that: $65536=2^{16}$ .

Hence, we need to find the sum of the prime factors of : $3^{16}-2^{16}$ .

In fact, the expression is factorisable since,

$3^{16}-2^{16}=(3^8+2^8)(3^8-2^8)$

$=(3^8+2^8)(3^4+2^4)(3^4-2^4)$

$=(3^8+2^8)(3^4+2^4)(3^2+2^2)(3^2-2^2)$

$=(3^8+2^8)(3^4+2^4)(3^2+2^2)(3+2)(3-2)$

$=(6817)(97)(13)(5)(1)$

Now, we need to check if $6817$ is prime or not as that's the only thing to be worried about because all the other are primes.

One can notice that $17\mid 6817$ by seeing $68$ and $17$ in $6817$ and hence $6817$ isn't a prime.

The following part is for those who didn't notice the above thing.

We note, $\sqrt{6817} \approx 82.5$ .

So, we need to check for odd integers(Since $6817$ is odd) less than $82$ which may divide $6817$ .

Our luck is quite good as we don't have to go much far as $17$ divides $6817$ and hence $6817$ is not a prime.

Now, dividing $6817$ by $17$ we get $401$ .

We check $401$ in a similar way if it's a prime or not.On checking, we get that it's a prime.

So,

$3^{16}-65536=3^{16}-2^{16}=(6817)(97)(13)(5)=(401)(17)(97)(13)(5)$ .

Hence,our required sum is: $401+17+97+13+5=\boxed{533}.$

First, we note that $65536 = 2^{16}$ .

Next, we use the identity $a^{2} - b^{2} = (a+b)(a-b)$ to factorize the given expression repeatedly:

$\begin{aligned}3^{16} - 2^{16} &= (3^{8} + 2^{8})(3^{8} - 2^{8})\\&= (3^{8} + 2^{8})(3^{4} + 2^{4})(3^{4} - 2^{4}) \\ &= (3^{8} + 2^{8})(3^{4} + 2^{4})(3^{2}+ 2^{2})(3^{2}-2^{2}) \\&= (3^{8} + 2^{8})(3^{4} + 2^{4})(3^{2}+ 2^{2})(3 + 2)(3 - 2)\end{aligned}$

$\begin{aligned}3^{16} - 2^{16} &= 6817 \times 97 \times 13 \times 5 \times 1\\ &= 17 \times 401 \times 97 \times 13 \times 5\end{aligned}$

We note that these factors are all prime and so:

$17 + 401 + 97 + 13 + 5 = \boxed{533}$

Notice that $65536=2^{16}$ . $\therefore 3^{16}-2^{16}$ $3^{16}-2^{16}=\left ( 3^{8} \right )^{2}-\left ( 2^{8} \right )^{2}$ $3^{16}-2^{16}=\left ( 3^{8}+2^{8} \right ) \left ( 3^{8}-2^{8} \right )$ Notice that $\left ( 3^{8}-2^{8} \right )$ can be further factorized by repeating the process where $a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right )$ . Hence, we will eventually obtain:

$\left ( 3^{16}-2^{16} \right )\equiv \left ( 3^{8}+2^{8} \right )\left ( 3^{4}+2^{4} \right )\left ( 3^{2}+2^{2} \right )\left ( 3^{2}-2^{2} \right )$ $3^{16}-2^{16}=\left ( 6817 \right )\left ( 97 \right )\left ( 13 \right )\left ( 5 \right )$

Because 6817 is not a prime number, hence we have to prime factorize $6817$ into $6817=17\times 401$ .

$\therefore 3^{16}-2^{16}=\left ( 17 \right )\left ( 401 \right )\left ( 97 \right )\left ( 13 \right )\left ( 5 \right )$

Thus, the sum of prime factors are $17+401+97+13+5=\boxed{533}$ .

(3^16) = 43046721

thus, (3^16) - 65536 = 42981185

leave the rest up to c++ ;-)

(write the include iostream statement here, I can't because the site treats it as a formatting character)

using namespace std;

bool isPrime (const int& number) {

bool property (true);

if (number % 2 == 0) { property = false; }

else {

for (int i = 3; i < number; i += 2) {

  if (number % i == 0) {

property = false;

break;

  }

}

}

return (property);

}

int main () {

unsigned int axa = 42981185;

for (int i = 2; i < axa; i++) {

if (axa % i == 0 && (isPrime (i)) == true) {

  cout << i << ", ";

}

}

cout << endl;

return (0);

}

3^16 - 65,536 = 42981185. The prime factorization of that number is 5 x 13 x 17 x 97 x 401. The sum of these numbers is 533. Therefore, 533 is the sum of the prime factors of 3^16 - 65,536.

First of all, note that $65536=2^{16}$ . Therefore, considering that $a^{2}-b^{2}=(a-b)(a+b)$ , we get that

$3^{16}-65536=3^{16}-2^{16}=(3^{8}+2^{8})(3^{8}-2^{8})=(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})=$ $=(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})=(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{1}+2^{1})(3^{1}-2^{1})$ .

Let us look at the brackets.

$3^{8}+2^{8}=6817=17\cdot401$ , which are both prime. $3^{4}+2^{4}=97$ , which is a prime. $3^{2}+2^{2}$ =13, which is a prime. $3+2=5$ , which is also a prime.

Thus, the sum we need is $17+401+97+13+5=\boxed{533}$

Only use calculator and the web http://www.mathsisfun.com/prime-factorization-tool.php :P

$65536=256^2$ $(3^8)^2-256^2=(3^8-256)(3^8+256)=((3^4)^2-16^2)(3^8+256)=(3^4-16)(3^4+16)(3^8+256)=65\cdot 97 \cdot 6817=5 \cdot 13 \cdot 97 \cdot 17 \cdot 401$ so the sum is equal to 533.

The problem above have solution 5 x 13 x 17 x 97 x 401 . So, 5 + 17 + 13 + 97 + 401 = 533 (ANSWERED)