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
·
3 years, 7 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'
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Vikram Waradpande
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3 years, 7 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.
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Adeeb Zaman
·
3 years, 7 months ago

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TopNewestAnother 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 · 3 years, 7 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 · 3 years, 7 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 · 3 years, 7 months ago

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– Takeda Shigenori · 3 years, 7 months ago

That is why i divided \( P-2 \) , or else it is always compositeLog in to reply