Number Theory: A problem of powers.

Question

I want you to check my proof:

$(\forall (x,y) \in \mathbb{N}^2)$ prove that if $2^x = 3y+1$ then $x$ is even and if $2^x = 3y+2$ then $x$ is odd.

$\underline{\textbf{Proof}}$: To solve this problem, we will put 2 cases, one in which $x$ is even and the other for $x$ as an odd number.

Let's go:

1. Case 1: Suppose $x$ is even so $x = 2a$ where $a \in \mathbb{N}$, so we can write that $2^x = 2^{2a} = 4^a$. And because $ 4 \equiv 1 \pmod 3$ then $4^a \equiv 1 \pmod 3$ and we remark that $4^a \not\equiv 2 \pmod 3$. And we're done for $x$ even.
2. Case 2: Suppose now $x$ is odd. To solve this case we will use that fact that now $x=2a+1$ so $2^x = 2^{2a+1} = 4^a\times2$ and we know that $4^a \equiv 1 \pmod 3 \Longrightarrow 4^a\times2 \equiv 2 \pmod 3 \Longrightarrow 4^a\times2 \not\equiv 1 \pmod 3$. And we're done for $x$ odd.

Answer 1

HINT Use the fact that $2^{2k} \equiv 1 \bmod 3$ and $2^{2k + 1} \equiv 2 \bmod 3$ because $2^{2k} = 4^{k} \equiv 1^k \equiv 1 \bmod 3$

Answer 2

Hint: If $x$ is odd, $2^x$ is of the form $$ 4^{\lfloor{x/2}\rfloor} \cdot 2$$

What is $4^{\lfloor{x/2}\rfloor} mod\, 3$ ?

Answer 3

I can give you a hint, to prove the first part assume that $x$ is even and $2^x = 3*y + 2$ now since x is a multiple of 2 L.H.S is a multiple of 4 and rhs is not a mutiple of 4 so this is not possible. Rest follows.