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.

#NumberTheory #Sharky

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Comments

Hint: The polynomial can be factored.

See answer here.

Pi Han Goh - 5 years, 7 months ago

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 - 5 years, 7 months ago

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

Sharky Kesa - 5 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}`	$\boxed{123}$