Question

Please help!
Prove: If n is congruent to 1(mod 6), then n^2 + 2^n is composite. Please help!
Prove: If n is congruent to 1(mod 6), then n^2 + 2^n is composite. @Mathematics

Answer 1

maybe use a contradiction and show it is prime

Answer 2

if n is congruent to 1(mod 6) then n=6k+1

Answer 3

ya i got that, then what?

Answer 4

replace in here n^2 + 2^n

Answer 5

replace what with that?

Answer 6

6k+1

Answer 7

for n

Answer 8

so n^2+2^n congruent to 1(mod 6) ?

Answer 9

hes saying:\[(6k+1)^2+2^{6k+1}\]Look at this expression and see if we can come to the conclusion that its composite.

Answer 10

now show that Joe wrote is divisible by 3

Answer 11

and thus composite

Answer 12

I guess the original problem should have stated for n >1? since 1^2+2^1=3, which is prime.

Answer 13

ya for n>1

Answer 14

so (6k+1)^2+2^(6k+1) congruent to 1 (mod 3) ?

Answer 15

if n is congruent to 1w/c is (mod 6) then (n=6k+1)

Answer 16

show that \[(6k+1)^2+2^{6k+1}\] is divisible by 3

or it is congruent to 0 mod 3

Answer 17

induction is pretty quick for this

Answer 18

hmm ok ty