Find Pythagorean triplet if one of it's member is 15 with formula m2-1 2m m2 - 1

A Pythagorean triplet consists of three positive integers \(a\), \(b\), and \(c\) such that \(a^2 + b^2 = c^2\). To find a Pythagorean triplet where one of the members is 15, we use the formula involving two integers \(m\) and \(n\), typically chosen such that \(m > n\). The formulas for the triplet are:

- \(a = m^2 - n^2\)
- \(b = 2mn\)
- \(c = m^2 + n^2\)

Given one of the members is 15, we need to determine \(m\) and \(n\) such that one of the above formulas equals 15.

### Case 1: \(a = 15\)
\[ m^2 - n^2 = 15 \]

Let's find \(m\) and \(n\) that satisfy this:

1. \( (m+n)(m-n) = 15 \)

Considering factor pairs of 15:
- \( 15 = 15 \times 1 \)

  Solving for \(m\) and \(n\):
  \[
  m+n = 15 \quad \text{and} \quad m-n = 1
  \]
  Adding these equations:
  \[
  2m = 16 \implies m = 8
  \]
  Subtracting these equations:
  \[
  2n = 14 \implies n = 7
  \]

Plugging \(m = 8\) and \(n = 7\) into the formulas for \(b\) and \(c\):
- \( b = 2mn = 2 \times 8 \times 7 = 112 \)
- \( c = m^2 + n^2 = 8^2 + 7^2 = 64 + 49 = 113 \)

Thus, the triplet is \( (15, 112, 113) \).

### Case 2: \(b = 15\)
\[ 2mn = 15 \]

Since \(2mn\) should be an integer, \(m\) and \(n\) should be such that their product divided by 2 equals 15:
\[ mn = 7.5 \]

This is not possible with integers, so \( b \neq 15 \).

### Case 3: \(c = 15\)
\[ m^2 + n^2 = 15 \]

This implies \(m^2 + n^2\) must be a sum of squares equal to 15, which is not possible with integers.

### Conclusion
The only valid Pythagorean triplet with one member being 15 using integer values for \(m\) and \(n\) is \( (15, 112, 113) \).

Would you like more details on the calculations or have any questions?

Here are some questions you might want to ask:
1. Can you explain how to derive the formulas for a Pythagorean triplet?
2. How do you verify if a set of numbers forms a Pythagorean triplet?
3. Are there any other methods to find Pythagorean triplets?
4. What are some common Pythagorean triplets?
5. How do Pythagorean triplets apply to real-world problems?

**Tip:** When working with equations involving integers, it's helpful to consider factor pairs and check if they meet the given conditions.

Discover how to find a Pythagorean triplet where one member is 15 using the formula m^2 - n^2, 2m, m^2 - 1. Step-by-step solution provided, including detailed calculations and exploration of integer solutions for m and n. Suitable for high school students studying number theory and algebraic equations.

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