0
$\begingroup$

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

Fermat's little theorem could help.

$\endgroup$
2
  • $\begingroup$ $q=2$ is the only solution. $\endgroup$
    – Lucian
    Aug 12, 2015 at 14:11
  • $\begingroup$ @Lucian figure out that, but how can I proof that is the only one? $\endgroup$
    – scummy
    Aug 12, 2015 at 14:23

1 Answer 1

3
$\begingroup$

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 $x\equiv \cdots \mod 9$)

$\endgroup$
1
  • $\begingroup$ Very nice. In particular it is not used the fact that $p$ is a prime number. $\endgroup$
    – Crostul
    Aug 12, 2015 at 15:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .