Either $np$ is even, or $n^2 − p^2$ is a multiple of $8$.

Question

Let $n$ and $p$ be two integers. Show that either $np$ is even, or $n^2 − p^2$ is a multiple of $8$.

If either one of $n$ or $p$ is even then $np$ is even and we are done. So let both of them is odd.

Let $n = 2m+1$ and $p = 2k + 1$. Then $$n^2 − p^2 = 4m^2+4m-4k^2-4k = 4(m-k)(m+k+1)$$

Now if both $m$ and $k$ are odd or even then $8 \mid 4(m-k)$ and we are done. If one is odd and one even then $8 \mid 4(m+k+1)$ and we are done.

So either $np$ is even, or $n^2 − p^2$ is a multiple of $8$.

Is the proof correct?

Yes, you could say directly that either $m-k$ or $m+k+1$ is even. — Thomas Lesgourgues, Jan 29 '19 at 15:58
Yes, but $p$ is a poor choice for an integer name. Usually it is a prime in this context. — Dietrich Burde, Jan 29 '19 at 15:59
Yes, the proof is correct (essentially via $\rm odd^2\equiv 1\pmod{8})$ and a dupe of many prior posts. — Bill Dubuque, Jan 29 '19 at 16:23

score 2 · Answer 1 · answered Jan 29 '19 at 16:02

2

Another approach from modular arithmetic: Notice if an integer $k$ is odd, then $k^2=1\text{ (mod } 8)$.

This follows from the fact that $1^2,3^2,5^2,7^2=1\text{ (mod } 8)$.

answered Jan 29 '19 at 16:02

Nick

1 Answers1