solutions of $p = 2q + r$

Question

let $P$ denote the rational primes of the for $4k+3$, and let $Q$ denote the set containing $1$ and all the rational primes of form $4k+1$. let $p \in P$. we look for representations of $p$ of the form $$ p = 2q + r $$ where $q,r \in Q$. for example: $$ 3 = 2 \cdot 1 + 1 \\ 7 = 2 \cdot 1 + 5\\ 11 = 2 \cdot 5 + 1 \\ 19 = 2 \cdot 1 + 17 \\ 23 = 2 \cdot 5 + 13 \\ 31 = 2 \cdot 1 + 29 \\ 43 = 2 \cdot 1 + 43 \\ 47 = 2 \cdot 17 + 13 \\ 59 = 2 \cdot 29 + 1 \\ 67 = 2 \cdot 13 + 41 \\ 71 = 2 \cdot 5 + 61 \\ 79 = 2 \cdot 13 + 53 \\ 83 = 2 \cdot 41 + 1 \\ 103 = 2 \cdot 1 + 101 \\ 107 = 2 \cdot 5 + 97 \\ 127 = 2 \cdot 13 + 101 \\ 131 = 2 \cdot 17 +97 \\ 139 = 2 \cdot 19 +101 \\ 151 = 2 \cdot 61 + 29 \\ 163 = 2 \cdot 61 + 43 \\ 167 = 2 \cdot 29 + 109 \\ 179 = 2 \cdot 53 + 73 \\ $$ question are these examples atypical, or are such representations fairly common, as appears from these first few values of $p$? i have no programming facilities at the moment, so, my curiosity having been caught, i would be grateful if someone could provide a reference to some relevant study and/or some estimates of the frequency with which a solution exists as $p$ increases in size.

Answer 1

It is conjectured that every odd number (greater $7$) can be written as $2q + r$ for odd primes $r,q$. This is called Lemoine's conjecture sometimes also Levy's conjecture and should be about as difficult as Goldbach's conjecture.

This is not exactly your question as you have an additional congruence condition; however there are variants of the Goldbach conjecture that also impose additional congruence conditions.

In any case, standard heuristics would predict that all but finitely many odd number congruent $3$ modulo $4$ can be written as $2q+r$ for primes congruent $1$ modulo $4$. To prove it might be somewhat harder than Goldbach's conjecture.