Waste less time on Facebook — follow Brilliant.
×

So succinct. It almost seems too easy

For all positive integers \(n>1\), prove that

\[n^5+n-1\]

has at least two distinct prime factors.

Note by Sharky Kesa
2 years 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

Hint: The polynomial can be factored.

See answer here.

Pi Han Goh - 2 years ago

Log in to reply

The factoring is really easy

\[n^5+n-1=(n^2-n+1)(n^3+n^2-1)\] (This could be manually done through comparing coefficients.)

The next thing to note is that they have to be both perfect powers to avoid having two distinct factors. We can easily check from here that they can't be perfect powers and thus they must have atleast two distinct factors.

I can post the proof on request but for now it is left as an excercise to the reader.

@Sharky Kesa now onto your functional equation.

Sualeh Asif - 2 years ago

Log in to reply

If you want, you can post the final part of the proof as a DM to me on Slack.

Sharky Kesa - 2 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...