Define the proposition P(n).

Let $P(n)$ be the statement $0^2 + 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$.

We check if $P(0)$ is true.
The Left Hand Side (LHS) is $0^2 = 0$.
The Right Hand Side (RHS) is $\frac{0(0+1)(2(0)+1)}{6} = \frac{0 \cdot 1 \cdot 1}{6} = 0$.
Since LHS = RHS, $P(0)$ is true.

Assume $P(k)$ is true for some integer $k \ge 0$. That is, assume:

We need to show that $P(k+1)$ is true. This means we need to prove:

Start with the LHS of $P(k+1)$:

Using the inductive hypothesis, substitute the sum up to $k^2$:

Factor out the common term $(k+1)$:

Find a common denominator for the terms inside the parenthesis:

Factor the quadratic expression $2k^2+7k+6$. It factors as $(k+2)(2k+3)$.

This matches the RHS of $P(k+1)$.

Since $P(0)$ is true and $P(k) \Rightarrow P(k+1)$, by the principle of mathematical induction, $P(n)$ is true for all integers $n \ge 0$.
The statement is proven: $\boxed{0^2 + 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}}$.

** (first part): Show that $F_{n-1}F_{n+1} = F_n^2 + (-1)^n$ for $n \ge 1$.**
The Fibonacci numbers are defined by $F_0 = 0$, $F_1 = 1$, and $F_k = F_{k-1} + F_{k-2}$ for $k > 1$.
The first few Fibonacci numbers are $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, \ldots$.

Let $P(n)$ be the statement $F_{n-1}F_{n+1} = F_n^2 + (-1)^n$.

We check if $P(1)$ is true.
LHS: $F_{1-1}F_{1+1} = F_0F_2 = 0 \cdot 1 = 0$.
RHS: $F_1^2 + (-1)^1 = 1^2 - 1 = 1 - 1 = 0$.
Since LHS = RHS, $P(1)$ is true.

Assume $P(k)$ is true for some integer $k \ge 1$. That is, assume:

We need to show that $P(k+1)$ is true. This means we need to prove:

Start with the LHS of $P(k+1)$:

Using the Fibonacci recurrence relation $F_{k+2} = F_{k+1} + F_k$:

From the inductive hypothesis, we know $F_k^2 = F_{k-1}F_{k+1} - (-1)^k$. Substitute this into the expression:

Factor out $F_{k+1}$:

Using the Fibonacci recurrence relation $F_k + F_{k-1} = F_{k+1}$:

Since $-(-1)^k = (-1) \cdot (-1)^k = (-1)^{k+1}$:

This matches the RHS of $P(k+1)$.

Since $P(1)$ is true and $P(k) \Rightarrow P(k+1)$, by the principle of mathematical induction, $P(n)$ is true for all integers $n \ge 1$.
The statement is proven: $\boxed{F_{n-1}F_{n+1} = F_n^2 + (-1)^n}$.