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

xp\frac{x}{p} is irreducible as long as xpx ≠ p and xx is not a multiple of pp

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...)

A Former Brilliant Member - 3 months, 1 week ago

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Ok. What about @Mahdi Raza?

A Former Brilliant Member - 3 months, 1 week ago

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

A Former Brilliant Member - 3 months, 1 week ago

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

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member Yes, but he answered in the note.

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member @Yajat Shamji

Fun fact

I and Mahdi Raza answered at the same time

Check the time on both of our comments/notes

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member Yes, I checked - it was 11 day ago...

You both win?...

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member Mmmmmhmmm

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

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member Sure!

What about this?:

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

x+nx + n is an integer and pp is a prime.

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

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member

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

Case 1 α\alpha is even:

If p>2pp>2\Rightarrow p is odd

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

If p=2p=2

pα\Rightarrow p\alpha is divisible by 44

Case 2 α\alpha is odd:

If pp is odd

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

If p=2p=2

pα\Rightarrow p\alpha is 2,x+n=12,x+n=1 or otherwise composite

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

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

therefore no answer is possible.

Zakir Husain - 3 months, 1 week ago

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

Putting this in your first condition ppn+p=pnp(pn+1)=pnpp_n+p=p_n\Rightarrow p(p_n+1)=p_n p=pnpn+10<p<1\Rightarrow p=\frac{p_n}{p_n+1}\Rightarrow 0<p<1 therefore no such prime is possible

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Rule 11 of the challenge (albeit the only one) - make a note on it - I'd advise using both proofs...

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member I think @Zakir Husain beat you to it...

A Former Brilliant Member - 3 months, 1 week ago

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@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 - 3 months, 1 week ago

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@A Former Brilliant Member Heh...

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

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member I got it incorrect anyways. I attempted it 11 to 22 hours before...

A Former Brilliant Member - 3 months, 1 week ago

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

A Former Brilliant Member - 3 months, 1 week ago

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@A Former Brilliant Member No problem!

A Former Brilliant Member - 3 months, 1 week ago

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