Show that if $p$ is an odd prime then the even perfect number
$$ 2^{p−1}(2^p− 1) \equiv 1 \pmod 9 $$
The prime $p$ is given as odd, and so $p-1$ is even, which we can write as $2k$ for some integer $k$.
Numerical experiments suggest that $2^{p-1}$ should be congruent to 1, 4 or 7 modulo 9. We'll prove a more general version of this statement by induction.
Let $S(k)$ be the statement
$$ S(k) := \quad 2^{2k} \equiv 1 \lor 4 \lor 7 \pmod 9 $$
We need to prove the base case $S(1)$ and the inductive step $S(k) \implies S(k+1)$.
Base Case $S(1)$
The base case is
$$ S(1) := \quad 2^{2 \times 1} \equiv 1 \lor 4 \lor 7 \pmod 9 $$
Since $2^2 \equiv 4 \pmod 9$, the base case is true.
Induction Step $S(k) \implies S(k+1)$
We assume $S(k)$ and aim to show $S(k+1)$, which is
$$ S(k+1) := \quad 2^{2(k+1)} \equiv 1 \lor 4 \lor 7 \pmod 9 $$
We start with
$$ 2^{2(k+1)} \equiv 4 \times 2^{2k} \pmod 9 $$
Since $S(k)$ is true, we have $2^{2k} \equiv 1 \lor 4 \lor 7 \pmod 9$.
Let's consider each case in turn.
$$ 2^{2k} \equiv 1 \pmod 9 \implies 2^{2(k+1)} \equiv 4 \times 1 \equiv 4 \pmod 9 $$
$$ 2^{2k} \equiv 4 \pmod 9 \implies 2^{2(k+1)} \equiv 4 \times 4 \equiv 7 \pmod 9 $$
$$ 2^{2k} \equiv 7 \pmod 9 \implies 2^{2(k+1)} \equiv 4 \times 7 \equiv 1 \pmod 9 $$
And so the inductive step is true for each case.
We have shown by induction that for integer $k \ge 1$
$$ \boxed{ 2^{2k} \equiv 1 \lor 4 \lor 7 \pmod 9 } $$
We now consider $2^p-1$ as follows, with integer $k \ge 1$
$$ \begin{align} 2^{p-1} & \equiv 2^{2k} \pmod 9 \\ \\ 2^p & = 2 \times(2^{2k}) \pmod 9 \\ \\ 2^p -1 & \equiv 2 \times (2^{2k}) -1 \pmod 9 \end{align} $$
Now, we've established that $ 2^{2k} \equiv 1 \lor 4 \lor 7 \pmod 9 $, so we'll consider each case in turn.
Case $2^{2k} \equiv 1 \pmod 9$
$$ \begin {align} 2^p-1 & \equiv 2 \times (1) -1 \pmod 9 \\ \\ & \equiv 1 \pmod 9 \\ \\ 2^{p-1}(2^p-1) & \equiv 1 \times 1 \pmod 9 \\ \\ & \equiv 1 \pmod 9 \end{align} $$
Case $2^{2k} \equiv 4 \pmod 9$
$$ \begin {align} 2^p-1 & \equiv 2 \times (4) -1 \pmod 9 \\ \\ & \equiv 7 \pmod 9 \\ \\ 2^{p-1}(2^p-1) & \equiv 4 \times 7 \pmod 9 \\ \\ & \equiv 1 \pmod 9 \end{align} $$
Case $2^{2k} \equiv 7 \pmod 9$
$$ \begin {align} 2^p-1 & \equiv 2 \times (7) -1 \pmod 9 \\ \\ & \equiv 4 \pmod 9 \\ \\ 2^{p-1}(2^p-1) & \equiv 7 \times 4 \pmod 1 \\ \\ & \equiv 1 \pmod 9 \end{align} $$
We've shown that in every case $2^{p-1}(2^p-1) \equiv1 \pmod 9$
The author's solution is simpler, which we'll set out here for practice.
We note that $2 \equiv -1 \pmod 3$. Since $p$ is odd, $p-1$ is even, and so $2^{p-1} \equiv 1 \pmod 3$. That is, for some integer $k$, we have $2^{p-1} = 3k + 1$.
We also note that $2 \times 2^{p-1} -1 = 2^p - 1 = 6k + 2 -1 = 6k + 1$.
Combining the two expressions gives us $2^{p-1}(2^p-1) = (3k+1)(6k+1) = 18k^2 + 9k+1$. This immediately gives us $2^{p-1}(2^p-1) \equiv 1 \pmod 9$.