# Perfect square

Let $$n \geq 1$$ be a positive integer and $$p$$ a prime.

If $$p \vert (n^{3} - 1)$$ and $$n \vert (p-1)$$, prove that $$4p - 3$$ is a perfect square.

###### Source: Gaussianos Blog

Note by Jordi Bosch
3 years, 6 months ago

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First, $$p|(n^{3} - 1) \Longrightarrow p|(n - 1)(n^{2} + n + 1)$$, so either $$p|(n - 1)$$ or $$p|(n^{2} + n + 1)$$.

In the first case we would have $$p \le (n - 1)$$. But we are given that $$n|(p - 1) \Longrightarrow n \le (p - 1)$$, which coupled with $$p \le (n - 1)$$ implies that $$n \le (n - 2)$$, an impossibility.

Thus we are now working with the facts that $$p|(n^{2} + n + 1)$$ and $$n|(p - 1)$$. Let $$p - 1 = kn$$ for some positive integer $$k$$. Then $$p = kn + 1$$, and since $$p|((n + 1)*n + 1)$$ we have that $$k \le (n + 1)$$.

Next, look at the case where $$k \lt (n + 1)$$. Let $$k = n - r$$ for some non-negative integer $$r$$. Then

$$p = kn + 1 = (n - r)*n + 1 = (n^{2} + n + 1) - (r + 1)*n \Longrightarrow p|(r + 1)*n$$.

Since $$p$$ cannot divide $$n$$, we must have that $$p|(r + 1) \Longrightarrow p \le (r + 1)$$.

But as $$p = kn + 1$$ and $$r = n - k$$ this would mean that $$p = (kn + 1) \le (n - k) + 1 \Longrightarrow kn \le (n - k)$$, which is an impossibility, since $$k$$ is a positive integer.

Thus $$k = n + 1$$, and so $$p = (n + 1)*n + 1 \Longrightarrow 4p - 3 = 4*(n^{2} + n + 1) - 3 = 4n^{2} + 4n + 1 = (2n + 1)^{2}$$,

a perfect square.

- 3 years, 6 months ago

Nicely done Brian!

- 3 years, 6 months ago

Thanks, Jordi. :)

- 3 years, 6 months ago

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