If x^2≡1 (mod 8), then x is :

The correct answer is Option (2) → odd integer

$x^2≡1 (mod\, 8)$

if $x$ is odd $(x=2k+1)$

$x^2=(2k+1)^2=4k^2+4k+1$

$∴ 4k^2+4k+1≡1 (mod\, 8)$

Hence, $x$ is an odd integer.