If n=4k+3 does 8 divide n^2-1?

If n = 4k + 3 does 8 divide n^2-1 [evenly; i.e., no remainder]?

(n^2 - 1) / 8 = (n + 1)(n - 1) / 8

= (4k + 3 + 1)(4k + 3 - 1) / 8

= (4k + 4)(4k + 2) / 8

= 4(k + 1)(2)(2k + 1) / 8

= 8(k + 1)(2k + 1) / 8

= (k + 1)(2k + 1) with no remainder.

So the answer is YES.