Multiple Choice Question (MCQ)
n^2 1 is divisible by 8 if n is

×
an even integer

✓
an odd integer

×
a natural number

×
an integer
n^2 1 is divisible by 8 if n is
an even integer
an odd integer
a natural number
an integer
If n is an odd integer,
let n = 2k+1
then
n^21
=(2k+1)^21
=4k^2+4k
Divide 4k^2+4k by 8
\dfrac{4k^2+4k}{8} = \dfrac{k^2+k}{2}=\dfrac{k(k+1)}{2}
Since k and k+1 is consecutive two numbers, there must be one is even number, which means k(k+1) is divisible by 2
Therefore, n^21 is divisible by 8 if n is an odd number