Proving there are infinitely many primes

[SP90]

Homework Statement

Prove that their are infinitely many primes such that $\left(\frac{14}{p}\right)=1$

Homework Equations

The bracketed symbol is the legendre symbol (i.e. there are infinitely many primes such that 14 is a square modulo p)

The Attempt at a Solution

Well by quadratic reciprocity $\left(\frac{14}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{7}{p}\right)= (-1)^{\frac{p^{2}-1}{2}}(-1)^{\frac{p-1}{2}\frac{7-1}{2}} \left(\frac{p}{7}\right)$.

The first part, $(-1)^{\frac{p^{2}-1}{2}}$ is 1 if $p=1,7(mod 8)$ or -1 if $p=3,5(mod8)$.

But I'm not sure how I can use that fact on the remainder of the equation, which would require modulo 7.

 

[mighty2000]

I would approach this problem by first understanding the concept of quadratic reciprocity and how it relates to the legendre symbol. Then, I would use mathematical induction to prove that there are infinitely many primes such that \left(\frac{14}{p}\right)=1.

First, we can see that if p=7, then \left(\frac{14}{7}\right)=1 since 14 is a perfect square. Next, we can assume that there exists at least one prime number, say q, such that \left(\frac{14}{q}\right)=1. This means that 14 is a square modulo q.

Now, we need to show that there exists a prime number, say r, such that \left(\frac{14}{r}\right)=1. By quadratic reciprocity, we have \left(\frac{14}{r}\right)=\left(\frac{2}{r}\right)\left(\frac{7}{r}\right). Since we have assumed that \left(\frac{14}{q}\right)=1, we know that \left(\frac{2}{q}\right)=1. This means that \left(\frac{2}{r}\right)=1 if r=1,7(mod 8) or \left(\frac{2}{r}\right)=-1 if r=3,5(mod 8).

Next, we need to consider the value of \left(\frac{7}{r}\right). If \left(\frac{7}{r}\right)=1, then we have found another prime number, r, such that \left(\frac{14}{r}\right)=1 and our proof is complete. If \left(\frac{7}{r}\right)=-1, then we can use the fact that r=1,7(mod 8) or r=3,5(mod 8) to show that \left(\frac{7}{r}\right)=-1 if r=3,5(mod 8). This means that we need to find a prime number, say s, such that s=3,5(mod 8) and \left(\frac{s}{7}\right)=-1. By quadratic reciprocity, this is equivalent to finding a prime number, s, such that s=3,5(mod 8) and \left(\frac{7}{s}\right)=-1. But