I am stuck on the end part here (or maybe im not) with this proof
Prove n^2-1 is a multiple of 8 whenever n is an odd integer.
Heres what I have done
Let n = even integer
therefore n = 2K k integer (im still unfamiliar with getting the symbols on the forums)
(2k)^2 - 1 is a multiple of 8?
4k^2 - 1 is multiple of 8,
4k^2 = even. even - 1= odd so that doesnt work
so
Let n = odd integer
Therefor n = 2k + 1 k integer
(2k+1)^2 - 1 multiple of 8
4k^2 + 4k +1 - 1 is a multiple of 8
4k^2 + 4K is a multiple of 8.
4k(k+1) is a multiple of 8..
(k+1) is an integer.
4k(k+1) is a multiple of 8. QED
Is this correct, or am I missing any parts?