# 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
3 months, 1 week ago

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

- 3 months, 1 week ago

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

Yes, but he answered in the note.

- 3 months, 1 week ago

@Yajat Shamji

Fun fact

I and Mahdi Raza answered at the same time

Check the time on both of our comments/notes

- 3 months, 1 week ago

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

You both win?...

- 3 months, 1 week ago

Mmmmmhmmm

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

- 3 months, 1 week 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

- 3 months, 1 week 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 $x+n=1;p_n=\frac{1}{p}\Rightarrow p_n \cancel{\in} Z$

- 3 months, 1 week 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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

Heh...

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

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

- 3 months, 1 week ago

No problem!

- 3 months, 1 week ago
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