Question

Find two positive values, m and n, that satisfy the following conditions:


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?
The value of m > n > 0.

in sentences

Answer 1

this using 9th grade math

Answer 2

To satisfy the conditions of finding two positive values \( m \) and \( n \) that are coprime, with one being even and the other odd, we can choose \( m = 3 \) and \( n = 2 \). These values meet the requirement because \( 3 \) is odd, \( 2 \) is even, and they are coprime, sharing no common divisors other than \( 1 \).

Using these values in Euclid's formula, which states that for any coprime integers \( m \) and \( n \) where \( m > n > 0 \), the primitive Pythagorean triple can be generated as follows:

\[
a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2.
\]

Plugging in our values:

\[
a = 3^2 - 2^2 = 9 - 4 = 5,
\]
\[
b = 2 \cdot 3 \cdot 2 = 12,
\]
\[
c = 3^2 + 2^2 = 9 + 4 = 13.
\]

Thus, the primitive Pythagorean triple generated is \( (5, 12, 13) \).

If both values \( m \) and \( n \) are either both even or both odd, they cannot be coprime because even integers share \( 2 \) as a common factor, and odd integers share \( 1 \) but fail to satisfy the coprime condition. As a result, primitive Pythagorean triples cannot be generated using both even or both odd integers under Euclid's formula, as it requires one integer to be even and the other to be odd to ensure coprimality.