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If $x^2≡1 (mod\, 8)$, then x is : |
even integer odd integer prime number composite number |
odd integer |
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. |
