Proof by induction that $5|11^n-6$ for all positive integers $n$

Question

Prove by induction that $5|11^n-6$ for all positive integers $n$

Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).$ Thus

$$f(k+1) = 11\cdot 11^k-11^k+f(k) = 10\cdot 11^k+f(k). $$

$5|(10\cdot 11^k+f(k))$ by hypothesis thus the statement is true for $p(k+1)$. Hence it's true for $p(n)$.

Is the above correct?

Answer 1

There are small corrections to your post. Define $a_{n}=11^{n}-6$. Let $P(n)$ be a predicate defined by $$P(n):5\mid a_{n}.$$ You have shown that $P(1)$ is true. Assume that $P(k)$ is true for some $k\in \mathbb{Z}^{+}$. Then $a_{k}=5m$ for some $m\in \mathbb{Z}$. Notice that $$a_{k+1}=11^{k+1}-6=5(11m+12).$$ This shows that $a_{k+1}$ is a multiple of $5$, so $5\mid a_{k+1}$. You can conclude the rest on your own.

Answer 2

Let $x=11^n-6$ then $$11^{n+1}-6=11(x+6)-6=11x+60$$ and this later is divisible by $5$ if $x$ is.

Answer 3

Suppose $5\mid 11^n-6\implies 11^n-6=5k$ for some $k\in\mathbb{Z}$. \begin{align*} 11(11^n-6)&=11\cdot5k \\ 11^{n+1}-11\cdot 6+60&=11\cdot 5k+60 \\ 11^{n+1}-6&=11\cdot 5k+60 \\ 11^{n+1}-6&=5(11k+12) \end{align*}

$$11^{n+1}-6=5(11k+12)\implies 5\mid 11^{n+1}-6$$

So we have shown $$5\mid 11^n-6\implies 5\mid 11^{n+1}-6$$

Check $5\mid 11^1-6$, and conclude $\forall n\in\mathbb{N},5\mid 11^n-6$.