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

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!

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

Explore how to find a Pythagorean triplet (a, b, c) where a < b < c and b + c equals a squared. This solution involves algebraic manipulation of the Pythagorean theorem and quadratic equations, presenting a step-by-step guide to determine the triplet. Suitable for students in Grades 9-10.

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