To find two positive values \(m\) and \(n\) that satisfy the conditions given, let's take \(m = 5\) and \(n = 2\).
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Checking the conditions:
- \(m = 5\) and \(n = 2\) satisfy \(m > n > 0\).
- Numbers \(5\) and \(2\) are coprime since their greatest common divisor is \(1\) (i.e., \(\gcd(5, 2) = 1\)).
- One of the integers, \(n\), is even.
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Using Euclid's formula to generate a primitive Pythagorean triple: Euclid's formula states that for two coprime integers \(m\) and \(n\) with \(m > n > 0\), the triple \( (a, b, c) \) can be generated with: \[ a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2. \]
Plugging in our values:
- \(a = 5^2 - 2^2 = 25 - 4 = 21\)
- \(b = 2 \cdot 5 \cdot 2 = 20\)
- \(c = 5^2 + 2^2 = 25 + 4 = 29\)
Thus, the primitive Pythagorean triple generated is \( (21, 20, 29) \).
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Discussion on the parity of \(m\) and \(n\): If both \(m\) and \(n\) are even, they cannot be coprime because any even integer will share at least the factor of \(2\). Therefore, you cannot generate a primitive Pythagorean triple using even values for both \(m\) and \(n\).
If both \(m\) and \(n\) are odd, then they could potentially still be coprime. However, the resultant \(a\), \(b\), and \(c\) from Euclid's formula would yield the following:
- Both \(a\) and \(b\) would end up being even (since \(2mn\) is even), and consequently \(c\) would also be odd, which contradicts the definition of a primitive Pythagorean triple, where one leg must be even and the other leg odd.
In conclusion, to generate a primitive Pythagorean triple, at least one of \(m\) or \(n\) must be even, while it is not possible to generate such a triple using both even or both odd values for \(m\) and \(n\).