Question

Sa se arate ca

(1+i)²⁰⁰⁸ + (1-i)²⁰⁰⁸ ∈ N

Answer 1

[tex]i^{2}=-1 [(1+i)^{2} ]^{1004}=(1+2i+i^{2})^{1004}=2i^{1004}

2^{1004}*(i^{2})^{502} 2^{1004}*(-1)^{502}= 2^{1004[tex] ∈ N

(1-i)^{2008}= (-2i)^{1004}=(-2)^{1004}*(i^2)^{504} 4^{504}* 1^{251} [/tex]} [/tex] ∈ N

$2^{1004} + 4^{504}$ ∈ N