$7^{n+2}+1$ is divisible by 8

Question

$n$ is an odd integer.

Use induction to prove $7^{n+2}+1$ is divisible by 8

When I use mathematical induction I get $$7^{k+3}+1$$

then when I use the assumption $$7\times 7^{k+2}+1\\ 7\times (8m-1)+1\\ 56m-6$$

How can I complete the answer? What went wrong?

Answer 1

What is wrong is that this statement is not true.

Consider solving for $k$ such that $8|7^{k+2}+1$.

Then the above statement is equivalent to $$\bmod 8: (-1)^{k+2} \equiv -1$$ which means $k$ must be odd for this to hold, and it does not hold for even $k$.

Answer 2

Since you are establishing the conjecture for "odd integers" (I assume positive), you have two options:

1. Replace $n$ by $2n-1$ and go on with induction as usual.
2. Show that the conjecture is true for $n=1$. In the inductive step, assume that the conjecture is true for some $n=k$ and prove that it is true for $n=k+2$.

Answer 3

There is another way, using binomial : Writing $7= 8-1$ and applying binomial theorem we get, $$7^{2n+1} + 1 = 1 + \sum_{r=0}^{2n+1}\binom{2n+1}{r}8^r (-1)^{2n+1-r} = 1+ (-1)^{2n+1} + \sum_{r > 0} 8^r \binom{2n+1}{r} (-1)^{2n+1-r}.$$ Thus $8$ divides the RHS.