prove or disprove that 2^(2k)-1 is never prime, if k is greater than or equal to 2, and a positive integer?

Answer 1

Let integer k >= 2.

2^(2k) - 1 = (2^k)^2 - 1

2^(2k) - 1 = (2^k)^2 - 1^2

2^(2k) - 1 = (2^k + 1)(2^k - 1)

For k >=2, (2^k + 1) >= 5 and (2^k - 1) >= 3

Both of these are distinct integer factors not equal to 1.

Thus, 2^(2k) - 1 is not prime. QED

Answer 2

2^(2k) -1 = (2^k+1)(2^k -1). It's the difference of
two squares and if k >=2, 2^k -1 > 1.
So 2^(2k) -1 is never prime.

Answer 3

2^(2k)-1
= (2^k + 1)(2^k - 1)
Since 2^(2k)-1 can be factored into two integers, 2^(2k)-1 is never prime.

Answer 4

How do you realize lim_{ok->infty} tan(ok)/ok has no cut back? From some attempting out for okay to up 100000000, it form of feels limsup tan(ok)/ok is approximately a million, so heuristically i could assume the s(n) to be at genuinely the worst merely n, which does no longer provide convergence. yet this is a awful estimate provided that very few ok's have tan(ok) ~ ok---say at worst a million in the type a million to n? with something being at worst tan(ok) ~ a million---which might provide convergence. in case you have a fashion for computing limsup {ok->infty} tan(ok)/ok, that could desire to probable be amplified to make those assertions rigorous, hence my question above.

Answer 5

The only prime even number is 2.