Question
Begin the discussion by responding to the following prompt.
Find two positive values, m and n, that satisfy the following conditions:
The value of m > n > 0.
Both m and n are coprime.
Either m or n is an even integer.
Use Euclid's formula to generate a primitive Pythagorean triple using these values of m and n.
What do you think will happen if the values of m and n are both even or both odd?
Can a primitive Pythagorean triple still be generated using values that are both even or both odd?
Find two positive values, m and n, that satisfy the following conditions:
The value of m > n > 0.
Both m and n are coprime.
Either m or n is an even integer.
Use Euclid's formula to generate a primitive Pythagorean triple using these values of m and n.
What do you think will happen if the values of m and n are both even or both odd?
Can a primitive Pythagorean triple still be generated using values that are both even or both odd?
Answers
GPT-4o mini
To find two positive values \( m \) and \( n \) that satisfy the conditions \( m > n > 0 \), that are coprime (i.e., their greatest common divisor is 1), and that one of them is even, we can choose:
Let \( m = 3 \) and \( n = 2 \).
1. **Check conditions:**
- \( m > n \): \( 3 > 2 \) (True)
- \( n > 0 \): \( 2 > 0 \) (True)
- Both \( m \) and \( n \) are coprime: \( \text{gcd}(3, 2) = 1 \) (True)
- At least one of them is even: \( n = 2 \) is even (True)
Now we can use Euclid's formula, which states that
\[
a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2
\]
to generate a primitive Pythagorean triple.
2. **Calculate the values:**
- \( a = 3^2 - 2^2 = 9 - 4 = 5 \)
- \( b = 2 \times 3 \times 2 = 12 \)
- \( c = 3^2 + 2^2 = 9 + 4 = 13 \)
Therefore, the primitive Pythagorean triple generated by \( m = 3 \) and \( n = 2 \) is \( (5, 12, 13) \).
### Discussion on even and odd values for \( m \) and \( n \)
If both \( m \) and \( n \) are **even**, then they cannot be coprime because both would share at least 2 as a common divisor. Therefore, it will be impossible to generate a primitive Pythagorean triple.
If both \( m \) and \( n \) are **odd**, while it is possible for them to be coprime, they will not generate a primitive Pythagorean triple. In that case, substituting them into Euclid's formula leads to:
- \( m^2 - n^2 \) (odd - odd = even)
- \( 2mn \) (odd * odd = odd, multiplied by 2 = even)
- \( m^2 + n^2 \) (odd + odd = even)
This means the resulting triple would not satisfy the criterion of being primitive, as a primitive Pythagorean triple contains one even and one odd number (the hypotenuse being the only number that could possibly be even, depending on the parity of \( m \) and \( n \)). Thus, the constructed Pythagorean triple would either repeat a factor or not remain coprime.
In conclusion, primitive Pythagorean triples can only be generated with one even and one odd integer \( m \) and \( n \). If both integers are even or both are odd, the corresponding triple will not be primitive.
Let \( m = 3 \) and \( n = 2 \).
1. **Check conditions:**
- \( m > n \): \( 3 > 2 \) (True)
- \( n > 0 \): \( 2 > 0 \) (True)
- Both \( m \) and \( n \) are coprime: \( \text{gcd}(3, 2) = 1 \) (True)
- At least one of them is even: \( n = 2 \) is even (True)
Now we can use Euclid's formula, which states that
\[
a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2
\]
to generate a primitive Pythagorean triple.
2. **Calculate the values:**
- \( a = 3^2 - 2^2 = 9 - 4 = 5 \)
- \( b = 2 \times 3 \times 2 = 12 \)
- \( c = 3^2 + 2^2 = 9 + 4 = 13 \)
Therefore, the primitive Pythagorean triple generated by \( m = 3 \) and \( n = 2 \) is \( (5, 12, 13) \).
### Discussion on even and odd values for \( m \) and \( n \)
If both \( m \) and \( n \) are **even**, then they cannot be coprime because both would share at least 2 as a common divisor. Therefore, it will be impossible to generate a primitive Pythagorean triple.
If both \( m \) and \( n \) are **odd**, while it is possible for them to be coprime, they will not generate a primitive Pythagorean triple. In that case, substituting them into Euclid's formula leads to:
- \( m^2 - n^2 \) (odd - odd = even)
- \( 2mn \) (odd * odd = odd, multiplied by 2 = even)
- \( m^2 + n^2 \) (odd + odd = even)
This means the resulting triple would not satisfy the criterion of being primitive, as a primitive Pythagorean triple contains one even and one odd number (the hypotenuse being the only number that could possibly be even, depending on the parity of \( m \) and \( n \)). Thus, the constructed Pythagorean triple would either repeat a factor or not remain coprime.
In conclusion, primitive Pythagorean triples can only be generated with one even and one odd integer \( m \) and \( n \). If both integers are even or both are odd, the corresponding triple will not be primitive.