Proof-that-n-Z-n-2-1-0-mod-4-

Question Number 176279 by CrispyXYZ last updated on 15/Sep/22

Proof that :

∄ n \in Z, n^{2} + 1 \equiv 0 (\mod 4)

Answered by mahdipoor last updated on 15/Sep/22

I > n = 2 k + 1

\Rightarrow n^{2} + 1 = 4 (k^{2} + k) + 2 \overset{4}{\equiv} 2

I I > n = 2 k

\Rightarrow n^{2} + 1 = 4 k^{2} + 1 \overset{4}{\equiv} 1

I & I I \Rightarrow ∄ n \in Z, n^{2} + 1 \overset{4}{\equiv} 0