Prove the following by induction:
$$ 2^{2n+1} ≡ 9n^2− 3n + 2 \pmod {54} $$
where $n$ is a natural number.
Let $P(n)$ be the statement
$$ 2^{2n+1} ≡ 9n^2− 3n + 2 \pmod {54} $$
We need to prove the base case $P(0)$ and the inductive step $P(n) \implies P(n+1)$.
Base Case
The base case is $P(0)$ which is
$$ 2^{2(0)+1} ≡ 9(0)^2− 3(0) + 2 \pmod {54} $$
This reduces to $2 \equiv 2 \pmod {54}$, which is true.
Inductive Step
To prove $P(n) \implies P(n+1)$ we assume the induction hypothesis $P(n)$ and aim to show $P(n+1)$, which is
$$\begin{align} 2^{2(n+1)+1} & \equiv 9(n+1)^2− 3(n+1) + 2 \pmod {54} \\ \\ 2^{2n + 3} & \equiv 9 n^2 + 15 n + 8 \end{align}$$
Let's start with the $2^{2n + 3}$
$$\begin{align} 2^{2n + 3} & = 2^{2n + 1} \times 2^2 \\ \\ & \equiv (9n^2− 3n + 2) \times 4 \pmod {54} \quad \text{by the induction hypothesis} \\ \\ & \equiv 36 n^2 - 12 n + 8 \pmod{54} \\ \\ & \equiv 9 n^2 + 15 n + 8 + 27n(n - 1) \pmod {54} \\ \\ & \equiv 9 n^2 + 15 n + 8 + 54m \pmod {54} \quad \text{since }n(n-1)=2m \text{ for some }m \\ \\ & \equiv 9 n^2 + 15 n + 8 \pmod {54} \quad \text{since } 5m \equiv 0 \end{align}$$
And so we have shown the inductive step is true.
By induction we have shown that $ 2^{2n+1} ≡ 9n^2− 3n + 2 \pmod {54} $ is true for all natural $n$.