I'm studying for a final and was wondering if this proof is correct. I've seen several variants but they all seem less logical, so I want to make sure I'm not overlooking something.
The proof is essentially: Prove, using induction that $5 | 11^{n} - 6$, for all positive integers n
Here is my proof: Base case, n = 1
$11^1 - 6 = 5$ which is divisible by 5.
Inductive hypothesis:: Assume the proposition is true for some $k \in Z^{+}$, or 5|$11^k - 6$
Inductive step: Prove, using the assumption made in the previous step that $5 | 11^{k+1} - 6$ From our inductive hypothesis we know that there exists some positive integer a such that 5a = $11^{k} - 6$
$5 | 11^{k+1} - 6 $
$11^{k+1} - 6 $ = $11^k * 11^1 - 6$ but, we know that 5a = $11^k - 6$, so substituting, we get: $11*5a = 55a$, which is a factor of 5.
Thus, the proposition holds.
An easier solution for the induction step may be that if $11^k-6=5m$ for some integer $m$,
i.e. $11^k=5m+6$
then $11^{k+1}=11(5m+6)=55m+66 = 5(11m+12)+6$
i.e. $11^{k+1}-6=5(11m+12)$
or in other words, if $11^k-6$ is a multiple of $5$ then $11^{k+1}-6$ is also a multiple of $5$