0

I understand that $x=3, c=1$ is a solution. Are there any more natural number solutions? If so, how would I find them? If not, why not?

IAAW
  • 1,486
  • 1
    By Euler's theorem, $2^6\equiv 1\pmod {3^2}$, so with $2^3 \equiv -1 \pmod{3^2}$, $2^{6n+3}\equiv-1 \pmod{3^2}$. – peterwhy Jun 22 '22 at 01:30

1 Answers1

2

Using mod $9$, and you will have: $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 7, 2^5 = 5, 2^6 = 1, 2^7 = 2$. Thus the only solution is $x = 3k$ where $k$ is odd. So the solution set is: $S = \{3,9,15,21,...\}$.

Wang YeFei
  • 6,390