Let $n$ be an integer. Prove that if $2|(n^2-1)$ then $4|(n^2-1)$.

Question

Let $n$ be an integer. Prove that if $2|(n^2-1)$ then $4|(n^2-1)$.

I know that $n^2=2k$ for some integer $k$. Please help me continue.

Answer 1

$n^2-1=(n+1)(n-1)$. Either these are both even or both odd; however if they are both odd then $n^2-1$ is odd.

Answer 2

$n^2-1$ factors into $(n+1)(n-1)$. If $2$ divides $n^2-1$, since $2$ is prime, it has to divide either $n+1$ or $n-1$. However, if one of them is even, so is the other. From here, you should be able to conclude that $4$ divides $(n+1)(n-1)$.

Answer 3

$2k=n^2-1=(n+1)(n-1)$ so both factors have to be even. That means that $(n+1)=2k_1$ and $(n-1)=2k_2$,for some integers $k_1$ and $k_2$, hence $n^2-1=4k_3$

Answer 4

Since $2$ divides $n^2-1$, this means $n^2-1$ is even, which means $n^2$ is odd, which in turn means that $n$ is odd. Hence, $n=2k+1$. We then have $$n^2-1 = (n+1)(n-1) = 4k(k+1)$$ Hence, we see that in fact $8$ divides $n^2-1$, since $2$ always divides $k(k+1)$.