Waste less time on Facebook — follow Brilliant.
×

Prime Problem

Prove that for every prime \(p>7\), \(p^6 -1\) is divisible by \(504\).

Note by Sudipta Biswas
3 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. 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"
MathAppears 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

Sort by:

Top Newest

By Fermat's Little Theorem, \(7\mid p^6-1\). By Euler's theorem, \(9\mid p^6-1\) and \(p^4\equiv 1\pmod 8\). Since odd squares leave residue \(1\) modulo \(8\), we have \(p^6=p^4\cdot p^2\equiv 1\cdot 1=1\pmod 8\). Finally because \(7,8,9\) are pairwise coprime, their product \(504\mid p^6-1\).

Jubayer Nirjhor - 3 years, 6 months ago

Log in to reply

\(504=2^{3} \times 3^{2} \times 7\) and by checking we can show the 6th residues \(mod 7,8,9\) are either \(0,1\) but since \(p\) is prime \(p^{6}=1\) mod \(504\) (by the Chinese remainder theorem as \(7,8,9\) are coprime) and hence we are done

Daniel Remo - 3 years, 6 months ago

Log in to reply

What is Chinese remainder theorem.??? Can you give me a link to any note..??

Sudipta Biswas - 3 years, 6 months ago

Log in to reply

Upon a quick google, http://www.math.tamu.edu/~jon.pitts/courses/2005c/470/supplements/chinese.pdf looks like the best link to explain it to you

Daniel Remo - 3 years, 6 months ago

Log in to reply

@Daniel Remo Also here I realised you don't really need it: ignoring the last step we have that \(7,8,9 | p^{6}-1\) and since they are coprime we know then that \(( 7\times 8\times 9) | p^{6}-1\)

Daniel Remo - 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...