elementary number theory - If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?

All is in the title:

If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?

I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.

Answer

Hint:

$$k^4 - 1 = (k^2 - 1)(k^2 + 1) = (k^2 - 1)\Big((k^2 - 1) + 2\Big)$$

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ADDED per comment: So yes, we have that if $(k^2 - 1)$ is divisible by $8$, then

$$k^4 - 1 = (k^2 - 1)(k^2 + 1) = 8b(k^2 + 1)$$ for some integer $b$.

And now, if $k^2 - 1$ is divisible by 8, it is even, then so is $k^2 + 1$.

That is, $k^2 + 1 = (k^2 - 1) + 2 = 8b + 2$. So

$$(k^2 - 1)(k^2 + 1) = 8b(8b + 2) = 16b (4b + 1)$$