Always prime

Is there a prime $P$ such that for any prime $p$ , $\frac{ (P-1)^p -1 }{ P-2 }$ is always prime?

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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}`	$\boxed{123}$

Comments

I think there isnt.let P-2=x.so P-1=1+x.now,finding the binomial expansion of(1+x)^p and putting it in our expression,it is found that our expression is always divisible by p.

Adeeb Zaman - 7 years, 8 months ago

That is why i divided $P-2$ , or else it is always composite

Takeda Shigenori - 7 years, 8 months ago

There cannot be such a function which always gives a prime. Its a well known fact. I think you can find the proof in Titu Andreescu's '104 Number Theory Problems'

Vikram Waradpande - 7 years, 8 months ago

Another empirical relation i founded was that, 2pn+1=prime number or a number having its factors as prime,where p is a prime number,n is any odd natural number

Avishkar Rajeshirke - 7 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}`	$\boxed{123}$