$p^3 = 2009 + 47 * 2^q$ where p and q are primes

Question

Solve the ecuations $p^3 = 2009 + 47 * 2^q$, where $p$ and $q$ are primes.

Fermat's little theorem could help.

Answer 1

Hint : Assume that $q\geq 5$ then $q$ is either of the form $6k+1$ or $6k-1$, taking the equation modulo $9$ gives us : $$2^{q}\equiv 2 \pmod 9 \text{ or } 2^{q}\equiv 5 \pmod 9 $$ hence $p^3\equiv \{6,3\}\pmod 9$ , but both cases are incompatible with the fact that for every integer $n$ we have : $$n^3\equiv \{0,1,-1\} \pmod 9 $$

(Notation I used $x\equiv \{a,b,\cdots\} \mod 9$ to say that $x\equiv a\mod 9$ or $x\equiv b \mod 9$ or $x\equiv \cdots \mod 9$)