Prove infinitude of primes of form 4k+1 using properties of Legendre symbol (-1/p)

[Vespero]

Homework Statement

Show that there are infinitely many primes $4k+1$ using the properties of $\left(\frac{-1}{p}\right)$.

Homework Equations

$\left(\frac{-1}{p}\right) = \begin{cases} 1, & \text{if }p\equiv 1\ (mod\ 4), \\ -1, & \text{if }p\equiv 3\ (mod\ 4). \end{cases}$

$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$

The Attempt at a Solution

Similar to other proofs involving the infinitude of the primes, I assumed that there were only a finite number of primes of the form 4k+1, specifically $p_1, p_2, ..., p_k$. I then formed the number $(p_{1}p_{2}...p_{k})^{2} + 1$. This number is not divisible by any $p_{i}, 1 \leq i \leq k$, for otherwise we would have $p_{i}|1$. Now, this number has a prime factorization by the Fundamental Theorem of Arithmetic as $(p_{1}p_{2}...p_{k})^{2} + 1 = q_{1}q_{2}...q_{r}$. From this point on, I'm not sure where to go. Clearly, I need to incorporate the Legendre symbol, but I can't see where to apply its properties to arrive at a contradiction.

 

[mighty2000]

To continue, we will use the property of the Legendre symbol that states:

\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)

Using this property, we can rewrite the prime factorization of (p_{1}p_{2}...p_{k})^{2} + 1 as:

(p_{1}p_{2}...p_{k})^{2} + 1 = q_{1}q_{2}...q_{r} = \left(\frac{p_{1}p_{2}...p_{k}}{p_{1}}\right)\left(\frac{p_{1}p_{2}...p_{k}}{p_{2}}\right)...\left(\frac{p_{1}p_{2}...p_{k}}{p_{r}}\right)

Since we know that p_{1}, p_{2}, ..., p_{k} are all of the form 4k+1, we can use the property of the Legendre symbol stated in the homework equations to see that:

\left(\frac{p_{1}p_{2}...p_{k}}{p_{i}}\right) = \left(\frac{p_{1}}{p_{i}}\right)\left(\frac{p_{2}}{p_{i}}\right)...\left(\frac{p_{k}}{p_{i}}\right)

Since p_{i} is also of the form 4k+1, the Legendre symbol \left(\frac{p_{i}}{p_{i}}\right) = 1. Therefore, we have:

\left(\frac{p_{1}p_{2}...p_{k}}{p_{i}}\right) = \left(\frac{p_{1}}{p_{i}}\right)\left(\frac{p_{2}}{p_{i}}\right)...\left(\frac{p_{i-1}}{p_{i}}\right)\left(\frac{1}{p_{i}}\right)\left(\frac{p_{i+1}}{p_{i}}\right)...\left(\frac{p_{k}}{p_{i}}\right)

Since 1 is congruent to