verify whether $3^n$ is divisible by $7$. For $n\ge1$

Question

Is there any other method to verify whether $3^n \equiv 0 \pmod 7$ for all positive integers? By mathematical induction I know the base case which is "$n=1$" will be false hence statement is false. but is there any other method to show that apart from mathematical induction.

Answer 1

The factorization of $3^n$ is... well, $3^n$. No $7$'s whereabouts.

Answer 2

For fun:

Euclid's Lemma:

If $p$, prime , divides $ab$, then $p$ divides $a$ or $p$ divides $b.$

Prove by induction that $7\not|3^n.$

1) True for $n=1.$

2) Hypothesis: $7 \not| 3^n.$

3)Step:

Assume $7|3^{n+1} :$

Then $7$ divides $3×3^n.$

By Euclid's Lemma $7$ divides $3$ or $7$ divides $3^n,$

by hypothesis $7\not| 3^n.$

Since $7 \not | 3 ,$

$\rightarrow:$

$7 \not | 3^{n+1}$.

Answer 3

It's obiously false for every $n$ by FTA.

You can also see it in different ways as an exercise.

EG

Suppose $\exists \bar n \quad \text{s.t} \quad3^{\bar n} \equiv 0 \pmod 7$

thus it should be true $\forall n>\bar n$

but for FLT

$3^6 \equiv 1 \pmod 7\implies \forall m \quad 3^{6m} \equiv 1 \pmod 7$

which is in cntrast with the assumption.

EG2

$3^0 \equiv 1 \pmod 7$

$3^1 \equiv 3 \pmod 7$

$3^2 \equiv 9\equiv 2 \pmod 7$

$3^3 \equiv 2\cdot 3 \equiv 6 \pmod 7$

$3^4 \equiv 6\cdot 3\equiv 4 \pmod 7$

$3^5 \equiv 4\cdot 3\equiv 5 \pmod 7$

$3^6 \equiv 5\cdot 3\equiv 1 \pmod 7$

For periodicity $\nexists n \quad \text{s.t} \quad3^{n} \equiv 0 \pmod 7$

EG3

Suppose:

$3^n=7k_1$

since $3\nmid 7 \implies 3 \mid k_1\implies 3^{n-1}=7k_2$

and so on, since you arrive to: $3=7k_n\implies 1=7k_{n+1}$

which is false.