What is the sum of the prime factors of $2^{16}-1$?

Question

I know $2^{10}=1024$ and $2^6=64$, but it seems they are not very helpful in solving this problem. There must be a trick to solve the problem in an easy way.

What is the sum of the prime factors of $2^{16}-1$?

Answer 1

Using $a^2-1=(a-1)(a+1)$ we have:

$$2^{16}-1=(2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=15\cdot 17\cdot 257=3\cdot 5\cdot 17\cdot 257$$

So, the answer is $3+5+17+257=282$.

Answer 2

HINT:

$$a^{2^{n+1}}-1=(a^{2^n}+1)(a^{2^n}-1)$$

as $$(b^m)^n=b^{mn}$$

See Power of Power

Answer 3

Every student should know the powers of two up to $2^{20}$ like the back of his/her hand, or in the case of a student with no hands, like some other familiar body part that would serve my rhetorical purpose.

Therefore we have $$2^{16}-1=65536-1 = 65535$$

Now any schoolchild can tell you that the semiprime $771$ is a factor: $$65535= 771\cdot 85$$

The rest is trivial:

$$65535=771\cdot 85 = 3 \cdot 5 \cdot 17 \cdot 257$$

Get a stray toddler to do the addition for you, to arrive at the final answer of

$$\boxed{282}$$