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# help me factorize p^4-1 as far as possible

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if someone can help me solve this I would appreciate it.

zerq152  Feb 15, 2018
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$$p^4-1$$ is a difference of squares. To make this clearer, the original expression can be rewritten as $$\left(p^2\right)^2-1^2$$. Since the original expression is a difference of squares, it is possible to factor it via $$(p^2+1)(p^2-1)$$. Notice that one of the factors of the original expression is also a difference of squares, namely $$p^2-1$$, so it is possible to factor even further. Therefore, the furthest factored expression would be $$(p^2+1)(p+1)(p-1)$$