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

#NumberTheory #Bounding #PrimeNumbers

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 6 years, 7 months ago

Nicely done Brian!

Jordi Bosch - 6 years, 7 months ago

Thanks, Jordi. :)

Brian Charlesworth - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$