# Is $\frac{x}{p}$ irreducible when $x$ is any integer and $p$ is a prime?

$\frac{x}{p}$ is irreducible as long as $x ≠ p$ and $x$ is not a multiple of $p$

Note by A Former Brilliant Member
7 months ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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} $\boxed{123}$

Sort by:

@Yajat Shamji, I did my best. The simplest proof (I think...)

- 6 months, 4 weeks ago

I think that @Mahdi Raza won - check Sum of squares.

- 6 months, 4 weeks ago

Did you ask that question to be created as a note @Yajat Shamji?

- 6 months, 4 weeks ago

Yes, but he answered in the note.

- 6 months, 4 weeks ago

@Yajat Shamji

Fun fact

I and Mahdi Raza answered at the same time

Check the time on both of our comments/notes

- 6 months, 4 weeks ago

Yes, I checked - it was $1$ day ago...

You both win?...

- 6 months, 4 weeks ago

Mmmmmhmmm

How about giving us another challenge @Yajat Shamji...?

- 6 months, 4 weeks ago

Sure!

Question: If $x + n + p$ and $p(x + n)$ both equal a prime $p_n$, prove that $\frac{x + n}{p}$ is also the same prime $p_n$.

$x + n$ is an integer and $p$ is a prime.

This time, though, your challenge is with @Zakir Husain

- 6 months, 4 weeks ago

• The condition you said is never true:
Let $\alpha=x+n$

Case 1 $\alpha$ is even:

If $p>2\Rightarrow p$ is odd

$\Rightarrow p\alpha$ is even therefore it can never be prime

If $p=2$

$\Rightarrow p\alpha$ is divisible by $4$

Case 2 $\alpha$ is odd:

If $p$ is odd

$p+\alpha$ is even therefore it can never be prime

If $p=2$

$\Rightarrow p\alpha$ is $2,x+n=1$ or otherwise composite

$\therefore$ $x+n=1$ is the only possible value

Now for $x+n=1;p_n=\frac{1}{p}\Rightarrow p_n \cancel{\in} Z$

- 6 months, 4 weeks ago

Another proof let $\frac {x+n}{p}=p_n\Rightarrow x+n=pp_n$

Putting this in your first condition $pp_n+p=p_n\Rightarrow p(p_n+1)=p_n$ $\Rightarrow p=\frac{p_n}{p_n+1}\Rightarrow 0 therefore no such prime is possible

- 6 months, 4 weeks ago

Rule $1$ of the challenge (albeit the only one) - make a note on it - I'd advise using both proofs...

- 6 months, 4 weeks ago

I think @Zakir Husain beat you to it...

- 6 months, 4 weeks ago

Yup, but I am sorry that I didn't read this comment till upto now @Yajat Shamji!

- 6 months, 4 weeks ago

Heh...

It seems as if I have a challenge for you...!

- 6 months, 4 weeks ago

I got it incorrect anyways. I attempted it $1$ to $2$ hours before...

- 6 months, 4 weeks ago

Oh... Sorry to hear that @Yajat Shamji!

- 6 months, 4 weeks ago

No problem!

- 6 months, 4 weeks ago