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 \(a > 7\), 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 \(m > n > 0\).

Find a Pythagorean triplet (a, b, c) such that a < b < c and (b + c) = a^2, a > 7

Explore how to find a Pythagorean triplet (a, b, c) such that a < b < c, b + c = a^2, and a > 7. This problem uses algebraic manipulation of the Pythagorean theorem, with solutions including (9, 40, 41). Ideal for high school students in Grades 10-12.

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