Waste less time on Facebook — follow Brilliant.
×

A Strange Number Theory proof?

There exists positive integers \(n\) such that \(2n + 1\) and \(3n + 1\) are perfect squares. Prove that \(n\) is divisible by \(40\).

Note by Sharky Kesa
2 years, 4 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

If \(n\) is odd, then \(2n+1=2(2k+1)+1=4k+3\). But \(3\) is not a quadratic residue mod \(4\).

If \(n\equiv \{2,4,6\}\pmod{8}\), then \(3n+1\equiv\{7,5,3\}\not\equiv x^2\pmod{8}\). Therefore \(8\mid n\).

If \(n\equiv\{1,3\}\pmod{5}\), then \(2n+1\equiv\{3,2\}\not\equiv x^2\pmod{5}\).

If \(n\equiv\{2,4\}\pmod{5}\), then \(3n+1\equiv\{2,3\}\not\equiv y^2\pmod{5}\). So \(5\mid n\).

Mathh Mathh - 2 years, 4 months ago

Log in to reply

Actually \(n\) is divisible by 40 as well. Try it. I've a solution, but I'm awaiting others to post theirs. Meanwhile, I'd say that the problem would have been much accurate had you asked us to prove that \(n\) is divisible by 40 instead of 8.

Satyajit Mohanty - 2 years, 4 months ago

Log in to reply

You could just post it directly to me on Slack.

Sharky Kesa - 2 years, 4 months ago

Log in to reply

Bhai, can you come on Slack ?

Swapnil Das - 2 years, 4 months ago

Log in to reply

Sure. Give me a few moments.

Satyajit Mohanty - 2 years, 4 months ago

Log in to reply

@Satyajit Mohanty Of Course!

Swapnil Das - 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...