To find a Pythagorean triplet $(a, b, c)$ such that $a < b < c$ and $b + c = a^2$, we start with the properties of Pythagorean triplets, which satisfy the equation:

$a^2 + b^2 = c^2$

We can manipulate the condition $b + c = a^2$ to express $c$ in terms of $b$ and $a$:

$c = a^2 - b$

Substituting this into the Pythagorean equation gives:

$a^2 + b^2 = (a^2 - b)^2$

Expanding the right side:

$a^2 + b^2 = a^4 - 2a^2b + b^2$

Simplifying this, we cancel $b^2$ from both sides:

$a^2 = a^4 - 2a^2b$

Rearranging gives:

$a^4 - 2a^2b - a^2 = 0$

This is a quadratic equation in $b$:

$2a^2b = a^4 - a^2$ $b = \frac{a^4 - a^2}{2a^2} = \frac{a^2(a^2 - 1)}{2a^2} = \frac{a^2 - 1}{2}$

Next, we can find $c$:

$c = a^2 - b = a^2 - \frac{a^2 - 1}{2} = \frac{2a^2 - (a^2 - 1)}{2} = \frac{a^2 + 1}{2}$

Thus, we have:

$a = a, \quad b = \frac{a^2 - 1}{2}, \quad c = \frac{a^2 + 1}{2}$

For $b$ and $c$ to be integers, $a$ must be odd (because $a^2 - 1$ is even when $a$ is odd). Let's take $a = 3$:

1. $b = \frac{3^2 - 1}{2} = \frac{9 - 1}{2} = 4$
2. $c = \frac{3^2 + 1}{2} = \frac{9 + 1}{2} = 5$

This gives the triplet $(3, 4, 5)$. We can verify:

$b + c = 4 + 5 = 9 = 3^2 = a^2$

Thus, the Pythagorean triplet is $(3, 4, 5)$ where $3 < 4 < 5$ and $b + c = a^2$.

If you need further details or have any questions, feel free to ask! Here are some follow-up questions you might consider:

1. Can you explain how to derive other Pythagorean triplets?
2. What are some properties of Pythagorean triplets?
3. How can we generalize the condition $b + c = a^2$?
4. What are examples of even and odd Pythagorean triplets?
5. How are Pythagorean triplets used in real-world applications?

Tip: Always check if the conditions of the problem are satisfied after finding a solution!