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!