For every natural number $n$, $ 3^{3n} - 1$ is divisible by $26$.

Question

Use induction to prove that for every natural number $n$, $ 3^{3n} - 1$ is divisible by $26$.

I can see that for $n=1$, $ 3^{3} -1=26\cdot 1$. As for inductive step, assuming that the statement holds for $n=k$ ($3^{3k}-1 = 26k$), I want to show it for $n=k+1$ (that is, $3^{ 3(k+1)} -1=26(k+1)$). But how to proceed from here?

Answer 1

From the assumption that for a particular $k$ we have $3^{3k}-1$ is a multiple of $26$ (which does not mean $3^{3k}=26k$), we have to prove that $3^{3(k+1)}-1$ is a multiple of $26$.

Note that $3^{3k+3}=27\cdot 3^{3k}$. But by the induction assumption $3^{3k}-1=26q$ for some integer $q$.

So $27\cdot 3^{3k}-1=27(26q+1)-1=26(27q+1)$, and therefore $3^{3k+3}$ s a multiple of $26$.

Answer 2

We have to proof that to $n\in\mathbb{N}^*,26\mid3^{3n}-1$, using induction.

for $n=1$

$3^{3\cdot1}-1=27-1=26$ wich is true, assuming that is true for $n$ then lets proof it for $n+1$

$3^{3\cdot(n+1)}-1=3^{3n+3}-1=27\cdot3^{3n}-1=(26+1)3^{3n}-1=26\cdot3^{3n}+3^{3n}-1$ since we are asuming its truth for $n$, then $26\mid3^{3n}-1$ and since $26\mid26\cdot3^{3n}$ then $26\mid26\cdot3^{3n}+3^{3n}-1=3^{3(n+1)}-1$.