By considering N=16*p1^2*p2^2*p3^2*...pn^2 - 2

prove that there are infinitely many primes of the form 8k-1.

I can't seem to deal with the fact that N is even here.

Please help!

Printable View

- Aug 8th 2009, 08:31 AMCairoPrimes
By considering N=16*p1^2*p2^2*p3^2*...pn^2 - 2

prove that there are infinitely many primes of the form 8k-1.

I can't seem to deal with the fact that N is even here.

Please help! - Aug 8th 2009, 01:26 PMOpalg
The key point here is that 2 is a quadratic residue modulo an odd prime p if and only if p is congruent to 1 or 7 (mod 8). Therefore $\displaystyle (4p_1p_2\cdots p_n)^2 - 2$ cannot have any prime divisors of the form 8k+3 or 8k+5. It clearly is divisible by 2, but all its other prime divisors must be congruent to 1 or 7 (mod 8), and at least one of them must be congruent to 7 (mod 8) (otherwise their product would be congruent to 2 (mod 16)).