To find a Pythagorean triplet $(a, b, c)$ such that $a < b < c$ and $b + c = a^2$, with $a \geq 7$, we’ll need to explore values for $a$ and check if corresponding values for $b$ and $c$ satisfy both the Pythagorean triplet condition ($a^2 + b^2 = c^2$) and $b + c = a^2$.

Solution:

1. Given $b + c = a^2$.
2. A Pythagorean triplet satisfies $a^2 + b^2 = c^2$.
3. $a \geq 7$.

Let's start with $a = 7$ and check if we can find $b$ and $c$ that satisfy both conditions.

For $a = 7$:

1. Calculate $a^2$: $a^2 = 7^2 = 49$

2. Set up equations: $b + c = 49$ $a^2 + b^2 = c^2 \Rightarrow 49 + b^2 = c^2$

3. Substitute $c = 49 - b$: Substitute $c$ in the equation $49 + b^2 = c^2$ and solve for $b$.

After some calculations, we find:

• $(7, 24, 25)$ is a Pythagorean triplet and