How to factor expressions with large powers of 2?

Question

I need to find the sole prime factor above $1000$ of $2^{22}+1$. I am not supposed to expand the number into its actual form. I began by factoring it as a sum of $11th$ powers, but this gave only $2^2+1=5$ as a factor, and left along with it an unfriendly collection of powers of two. I am unsure how to proceed.

Answer 1

Write it as $4 \cdot 2^{20} + 1$ and use the factorization

$$4x^4 + 1 = (2x^2 + 2x + 1)(2x^2 - 2x + 1)$$

(a special case of the Sophie Germain identity) to get

$$4 \cdot 32^4 + 1 = (2 \cdot 32^2 + 2 \cdot 32 + 1)(2 \cdot 32^2 - 2 \cdot 32 + 1).$$

We have $2 \cdot 32^2 = 2048$ so the first factor is $2048 + 64 + 1 = 2113$ and the second factor is $2048 - 64 + 1 = 1985$. The latter is divisible by $5$ so the former must be the unique prime factor above $1000$, assuming that it exists (we could also check that it's prime).