Find a Pythagorean triplet (a, b,c) such that a < b < c and (b+c) = a*a, a>=7

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

Find a Pythagorean triplet (a, b, c) such that a < b < c and (b + c) = a * a, a >= 7

Explore how to find a Pythagorean triplet (a, b, c) satisfying a < b < c and b + c = a^2, with a starting from 7. This guide covers the Pythagorean theorem and provides a detailed solution. Ideal for high school students in Grades 9-12.

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