Is $2^{57} + 1$ is a composite number?

Question

Prove or disprove the following: $2^{57} + 1$ is a composite number.

@Elborito it doesn't matter because people will answer it anyway — Randall, Jan 25 '19 at 03:23
@Randall Yes I know, but she can write the context of the question, it is from of a book or a class or Algebra, etc. — Boris Valderrama, Jan 25 '19 at 03:33
https://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd — lab bhattacharjee, Jan 25 '19 at 03:52

score 5 · Answer 1 · answered Jan 25 '19 at 03:15

5

Yes: $2^{57}+1\equiv(-1)^{57}+1=-1+1=0\mod 3$.

answered Jan 25 '19 at 03:15

sranthrop

8,497

score 3 · Accepted Answer · edited Jan 25 '19 at 03:58

3

We can factor $$2^{57}+1 = \left( 2^{19} \right)^3+1 = \left( \left(2^{19} \right)^2 -2^{19} + 1 \right)\left( 2^{19} + 1 \right)$$ where both factors are greater than $1$. Therefore, $2^{57}+1$ is composite.

edited Jan 25 '19 at 03:58

Lee David Chung Lin

7,043

answered Jan 25 '19 at 03:22

JimmyK4542

54,331

Is $2^{57} + 1$ is a composite number?

2 Answers2