Question

Prove, if n=4k+3 for some integer k, then 8

dividesn2-1 (another way of showing
is8|n2-1).

Answer 1

See proof below

Explanation:

Let
`n=4k+3` for some integer k.

Multiply n by itself to get
`n^2=(4k+3)(4k+3)=[16k^2+12k+12k+9=16k^2+24k+9`

Now substract 1 in both sides of the equation, and factor 8 to get

`n^2-1=16k^2+24k+8=8(2k^2+3k+1)=8m`, if we define
`m=2k^2+3k+1`.

Thus,
`n^2-1=8m` for some integer m, that is,
`8|n^2+1`