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

@A Former Brilliant Member –

• The condition you said is never true:

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

therefore no answer is possible.

@A Former Brilliant Member – Another proof let $\frac {x+n}{p}=p_n\Rightarrow x+n=pp_n$

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

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

@A Former Brilliant Member – Oh... Sorry to hear that @Yajat Shamji!