Quadratic Reciprocity Question

I'm having trouble understanding the proof for 2 being a quadratic residue mod p <=> p = plus or minus 1 mod 8. I'm using the Springer Elementary Number Theory text. For (=>), the text says (p^2-1)/8 is even <=> 16 divides (p^2-1) <=> 16 divides (p+1)(p-1) <=> 8 divides (p+1) or 8 divides (p-1). My trouble is with this last line. I know I'm missing something obvious, but I don't see how 16 divides (p+1)(p-1) <=> 8 divides (p+1) or 8 divides (p-1). Can anyone help?

Re: Quadratic Reciprocity Question

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.

Re: Quadratic Reciprocity Question

Oh, I see now. I wasn't sure it was something you could tell just by looking at the bicondiontional, or if it was something you had to do a few test cases for. Thank you, this was a very clear explanation.