To find a Pythagorean triplet $(a, b, c)$ such that $a < b < c$ and $b + c = a^2$ with $a > 7$, we can start from the Pythagorean theorem, which states:

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

Given the condition $b + c = a^2$, we can express $c$ in terms of $b$ and $a$:

$c = a^2 - b$

Substituting $c$ in the Pythagorean theorem gives us:

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

Expanding and simplifying this equation:

$a^2 + b^2 = a^4 - 2a^2b + b^2$ $a^2 = a^4 - 2a^2b$ $0 = a^4 - 2a^2b - a^2$ $0 = a^4 - a^2 - 2a^2b$ $2a^2b = a^4 - a^2$ $b = \frac{a^4 - a^2}{2a^2} = \frac{a^2(a^2 - 1)}{2a^2} = \frac{a^2 - 1}{2}$

Now, substituting $b$ back to find $c$:

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

Now we have:

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

Next, we need to check specific values of $a > 7$:

1. For $a = 8$: $b = \frac{8^2 - 1}{2} = \frac{64 - 1}{2} = \frac{63}{2} = 31.5 \quad \text{(not an integer)}$

2. For $a = 9$: $b = \frac{9^2 - 1}{2} = \frac{81 - 1}{2} = \frac{80}{2} = 40$ $c = \frac{9^2 + 1}{2} = \frac{81 + 1}{2} = \frac{82}{2} = 41$ Here, $(a, b, c) = (9, 40, 41)$.

3. For $a = 10$: $b = \frac{10^2 - 1}{2} = \frac{100 - 1}{2} = \frac{99}{2} = 49.5 \quad \text{(not an integer)}$

Continuing this way, we can derive more triplets.

Thus, one valid triplet is $(9, 40, 41)$ where $9 < 40 < 41$ and $b + c = 40 + 41 = 81 = 9^2$.

Summary:

A Pythagorean triplet that meets your criteria is: $(a, b, c) = (9, 40, 41)$

Would you like further details or examples? Here are some related questions you might consider:

1. How do you prove that $(9, 40, 41)$ is indeed a Pythagorean triplet?
2. Can you find more Pythagorean triplets with $a > 7$?
3. What is the general formula for generating Pythagorean triplets?
4. How do you verify the condition $b + c = a^2$?
5. What are some applications of Pythagorean triplets in real life?

Tip: When generating Pythagorean triplets, consider using the formula $m^2 - n^2, 2mn, m^2 + n^2$ for integers $m > n > 0$.