Prove that if $n$ is an odd integer, then $n^2-1$ is divisible by $8$.

Question

I have it proved for when k is odd, but when k is even, I get 8[q(2q+1)]. Is it correct to state that 8(Integer) proves that n^2-1 is divisible by 8?

Answer 1

There are two cases, $k$ is odd so $k+1$ is even or $k$ is even so $k+1$ is odd. In either case we have one is divisible by $2$.

Answer 2

Either $k$ or $k+1$ is divisible by 2. Then $k(k+1)$ is divisible by 2 and you're done.

Answer 3

Hint: Either $k$ or $k+1$ is even.

Answer 4

$$n^2-1=(n+1)*(n-1)$$ If $n\equiv3\mod4$, then $n+1\equiv0\mod4$ and $n-1\equiv2\mod4$.

If $n\equiv1\mod4$, then $n+1\equiv2\mod4$ and $n-1\equiv0\mod4$.

Either way, we have $(4k)*(4k+2)=16k^2+8k\equiv0\mod8$.

Answer 5

Proof by induction:

This is true for $n=1 (0)$ and $n=3 (8)$.

For the induction step,

$\begin{array}\\ ((2n+3)^2-1)-((2n+1)^2-1) &=(4n^2+12n+9-1)-(4n^2+4n+1-1)\\ &=(12n+8)-(4n)\\ &=8n+8\\ \end{array} $

which is obviously divisible by $8$.