Primes of the form $5n+4$.

Question

Is there an elementary way of showing the infinitude of primes of the form $5n+4$ without using quadratic reciprocity?

Answer 1

I wanted to give this as a comment but it was too long. For a proof one must consider following points:

*n must be odd and $5n+4$ results a set of infinite numbers of form $...9=10 x +9$:

$9, 19, 29, 39, 49, 59, 69, 79, . . .$

• This set has common members with the set of primes resulted by known generating functions,in fact if n is large enough $5n+4$ can always be transformed to a polynomial identical to one form of generating function, here is some example:

$79=2\times 3\times 13 +1$ (Euclidean theorem for primes)

$59=2\times 29+1$, $ 19=4\times 5 -1 $, $149=6\times 25 -1$, represent samples of Dirichlet primes resulted from $2x+1$, $4x-1$ and $6x-1$ respectively.

$229=2\times 10^2 +29$ represent a sample of $2x^2+29$ which is Euler's function.

$1129=30\times 37 + 19$ which represent a theorem that says all primes are of form $30 x+r$ where $r=1, 7, 11, 13, 19,23, 29 $ particularly when $r=19$ and $r=29$; in other words primes resulted from $5n+4$ are members of a subset of primes of form $30 x +19$ or $30 x+29$.