Number Systems
$$n^2 -1$$ is divisible by $$8$$ , if n is
If $$n$$ is an odd integer then show that $${n^2} - 1$$ is divisible by 8
Any odd positive integer is in the form of 4p+1 & 4p+3 for n=4p+1
$$(n^{2}-1)=(1p+1)^{2}-1=16p^{2}+8p+1=16p^{2}+8p=8p(2p+1)$$
$$=(n^{2}-1)$$ is divisible by 8.
$$(n^{2}-1)=(4p+3)^{2}-1=8(2p^{2}+3p+1)$$
$$\Rightarrow n^{2}-1$$ is divisible by 8
$$n^{2}-1$$ is divisible by 8
$$n^2 -1$$ is divisible by $$8$$ , if n is
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