$P^{2} - 1$ is always divisible by (if P is an odd positive integer)

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Solution

The correct options are
A 4
D 8
Every odd number can be written in the form of $4 k + 1$ or $4 k + 3$ where $k$ is an integer.
Case i) $p = 4 k + 1$
Then $p^{2} - 1 = 16 k^{2} + 8 k = 8 (2 k^{2} + k)$
We see that $p^{2} - 1$ is divisible by $8$ and hence also by $4$ .
Case ii) $p = 4 k + 3$
Then $p^{2} - 1 = 16 k^{2} + 24 k + 8 = 8 (2 k^{2} + 3 k + 1)$
We see that $p^{2} - 1$ is divisible by $8$ and hence also by $4$ .
So in both the cases $p^{2} - 1$ is divisible by $8$

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