well if 16 divides either p-1 or p+1 we're done. and if 2 divides p-1 but 4 does not, then 8 must divide p+1 (since 16 divides (p-1)(p+1)).

similarly if 2 divides p+1 but 4 does not, 8 must divide p-1.

which leaves us with one possibility left: 4 divides p-1, but 8 does not, and 4 divides p+1 but 8 does not.

but if 4 divides p-1, say p-1 = 4k, and 8 does not, k must be odd. so p-1 = 4(2t+1) for some integer t.

then p+1 = (p-1) + 2 = 4(2t+1) + 2 = 8t + 4 = 8t + 6. so p = 8t + 5, and (p^{2}-1)/8 = (64t^{2}+ 80t + 24)/8 = 8t^{2}+ 10t + 3, which is odd, contradicting the hypothesis that (p^{2}- 1)/8 is even.

another way to look at it: if p-1 = 4k and p+1 = 4k+2 where k is odd, then:

p^{2}- 1 = (4k)(4k+2) = 16k^{2}+ 8k = 8k(2k+1) and 16 does not divide this.