<s>[INST] Mathematical induction proof $n^4-1$ is divisible by $16$ for all odd integers $n$ I'm stuck towards the end of proving this, here's my attempt:
$P(3) = 80/16 = 5$, True
$P(k) = k^4-1$
$P(k+1)= (k+1)^4-1$
Expanded $= k^4+4k^3+6k^2+4k+1-1$
This is where I am stuck at.
Sorry for the sloppy formatting im still reading how to  properly write formulas on this site.
Thanks in advance for your help!
 [/INST] Without loss of generality we proved the statement for positive odd integers.
Note that a positive odd integer has the form of $2k-1$ where $k\ge 1$ 
We proceed with  induction on k.
Let $P(k)$ be  $(2k-1)^4-1$ is a multiple of $16$
For $k=1$, $(2k-1)^4-1=0$ which is a multiple of $16$ so $P(1)$ is true.
Assume that $(2k-1)^4-1$ is a multiple of $16$
We need to show that $(2k+1)^4-1$ is also a multiple of $16$
Note that the binomial theorem implies $$(2k+1)^4 = [(2k-1)+2]^4 = (2k-1)^4+8(2k-1)^3 +24(2k-1)^2+32(2k-1)$$
Thus $$(2k+1)^4 - (2k-1)^4 =8(2k-1)^3 +24(2k-1)^2+32(2k-1)=$$
$$ 8(2k-1)[(2k-1)^2+3(2k-1)+4)]$$
Note that the expression in square bracket is an even integer, so the result is multiple of $16$ 
</s>