Proof that $(3\cdot 2^n-1)$ is not a multiple of $17$ for any value of $n$

Question

Prove that $3\cdot 2^n-1$ is not a multiple of $17$ for any positive integer $n$.

Answer 1

For $n=4k$, $k \geq 0$ we have that $$2^n \equiv \pm1 \pmod {17} \Rightarrow 3\cdot 2^n-1 \equiv 2,-4 \equiv 2,13 \pmod {17}$$ For $n=4k+1$, $k \geq 0$ we have that $$2^n \equiv \pm2 \pmod {17} \Rightarrow 3\cdot 2^n-1 \equiv 5,-7 \equiv 5,10 \pmod {17}$$ For $n=4k+2$, $k \geq 0$ we have that $$2^n \equiv \pm4 \pmod {17} \Rightarrow 3\cdot 2^n-1 \equiv 11,-13 \equiv 11,4 \pmod {17}$$ For $n=4k+3$, $k \geq 0$ we have that $$2^n \equiv \pm8 \pmod {17} \Rightarrow 3\cdot 2^n-1 \equiv 23,-25 \equiv 6,9 \pmod {17}$$

Hence we can conclude that $\forall$ $n \in \mathbb{N}$, $17$ does not divide $(3\cdot 2^n -1)$.

Answer 2

$$2^n \equiv 1,2,4,8,16,15,13,9 \pmod {17}$$ $$3 \cdot 2^n \equiv 3,6,12,7,14,11,5,10 \pmod {17} $$

Answer 3

Suppose there is $n$ such $$ 3 \cdot 2^{n} \equiv 1 \pmod{17}. $$ Multiply by 6 on both sides and simplify by $2$ (which is coprime to $17$) to get $$ 2^{n-1} \equiv 3 \pmod{17}. $$ But $$ 2^{8} = (2^{4})^{2} \equiv (-1)^{2} \equiv 1 \pmod{17}, $$ while $$ 3^{8} = (3^{4})^{2} \equiv (-4)^{2} \equiv -1 \pmod{17}. $$

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EDIT after a suggestion of @fleablood to spell out the ending.

This leads to the contradiction $$ 1 \equiv (2^{8})^{n-1} \equiv (2^{n-1})^{8} \equiv 3^{8} \equiv -1 \pmod{17}. $$