Which Prime Divisors of 4n^2+4n-1 Are Congruent Modulo 8 to \pm 1?

[TTob]

I need to prove that p congruent modulo 8 to \pm 1 for every prime divisor p of 4n^2+4n-1.
4n^2+4n-1 is odd so we have
p \equiv \pm 1,3,5 \pmod{8}

I don't know how to continue from here... I need some hint.

Thanks.

 

[Martin Rattigan]

I need to prove that p congruent to \pm 1 for every prime divisor p of 4n^2+4n-1.

Congruent modulo what? 7|(4.5^2+4.5-1), but 7\neq\pm 1(5), so presumably the answer is not n.

 

[TTob]

I need to prove that p congruent modulo 8 to \pm 1.

 

[Martin Rattigan]

4n^2+4n-1=(2n+1)^2-2=0(p) only if 2 is a quadratic residue mod p which is true only if p=\pm 1(8). (This is proved in any number theory text in the section on quadratic reciprocity.)

 

[TTob]

4n^2+4n-1=(2n+1)^2-2=0(p) only if 2 is a quadratic residue mod p which is true only if p=\pm 1(8). (This is proved in any number theory text in the section on quadratic reciprocity.)

Thank you !

 

[robert Ihnot]

p \equiv \pm 1,3,5 \pmod{8}

Boy! that was a confusing bunch of stuff. Lot of times, the trouble with a problem is the writer did not correctly organize what he wants to solve. But I see, TTob got it all correct!