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Question

If pbe a natural number, then prove that $$p^{n \, + \, 1} \, + \, (p \, + \, 1)^{2n \, - \, 1} $$ is divisible by  $$p^2$$ +p + 1 for every positive integer n.


Solution

Let f (n) = $$p^{p \, + \, 1} \, + \, (p \, + \, 1)^{2n \, - \, 1}$$
We have f (1) = $$p^2$$ + p + 1 so that f (1) is divisible by $$p^2$$ + p + 1.
Now assume that f (m) is divisible by $$p^2$$ + p + p + 1 i.e.,
we assume that
$$p^{m \, + \, 1} \, + \, (p \, + \, 1)^{2m \, - \, 1} \, = \, k(p^2 \, + \, p \, + \,1)$$
Now f (m + 1) = $$p^{m \, + \, 2} \, + \, p \, (p \, + \, 1)^{2m \, + \, 2 \, - \, 1}$$
$$= \, p^{m \, + \, 2} \, + \, +[k(p^2 \, + \, p \, + \, 1) \, - \, p^{m \, + \, 1}](p \, + \, 1)^2$$
$$= \, p^{m \, + \, 2} \, - \, (p \, + \, 1)^2 \, p^{m \, + \, 1} \, + \, k(p \, + \, 1)^2 \, (p^2 \, + \, p \, + \, 1)$$
$$= \, p^{m \, + \, 1} \, (p \, - \, p^2 \, - \, 2p \, - \, 1) \, + \, k(p \, + \, 1)^2 \, (p^2 \, + \, p \, + , 1)$$
$$= \, (p^2 \, + \, p \, + \, 1) \, [k(p \, + \, 1)^2 \, - \, p^{m \, + \, 1}]$$
Hence f (m + 1) is divisible by $$p^2$$ + p + 1.
$$\therefore$$ By induction, f (x) is divisible by $$p^2$$ + p + 1 for all n $$\epsilon$$ N.

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