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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, 2 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.

Brian Charlesworth - 3 years, 2 months ago

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Nicely done Brian!

Jordi Bosch - 3 years, 2 months ago

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Thanks, Jordi. :)

Brian Charlesworth - 3 years, 2 months ago

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