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Always prime

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

Note by Takeda Shigenori
4 years, 3 months ago

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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 - 4 years, 3 months ago

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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 - 4 years, 3 months ago

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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 - 4 years, 3 months ago

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That is why i divided \( P-2 \) , or else it is always composite

Takeda Shigenori - 4 years, 3 months ago

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