<s>[INST] 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 $16$.
 [/INST] $\displaystyle (2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1=8b+1$ where $b$ is some integer
Now, $(2a+1)^4=(8b+1)^2=64b^2+16b+1\equiv1\pmod{16}$
This way we can show $(2a+1)^{2^n}\equiv1\pmod{2^{n+2}}$ for integer $n\ge1$
This is generalized as  Carmichael function $$\lambda(2^{n+2})=2^n\text{ for }n\ge1$$
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