Is it possible to write a Pythagorean triple in which 2 of the numbers are even and one is odd? Explain your reasoning.
To determine whether it is possible to write a Pythagorean triple with two even numbers and one odd number, we can use the properties of Pythagorean triples.
A Pythagorean triple consists of three positive integers a, b, and c, such that a^2 + b^2 = c^2. The numbers a, b, and c represent the lengths of the sides of a right-angled triangle, with c being the hypotenuse.
Let's consider an even number. Any even number can be written as 2n, where n is a positive integer. Similarly, any odd number can be written as 2m + 1, where m is a positive integer.
Now, let's substitute these expressions into the Pythagorean equation:
(2n)^2 + (2m + 1)^2 = c^2
Expanding the equation, we get:
4n^2 + 4m^2 + 4m + 1 = c^2
We can simplify this equation by dividing both sides by 4:
n^2 + m^2 + m + 1/4 = c^2/4
Since n^2 and m^2 are perfect squares, c^2/4 must also be a perfect square. In other words, c must be an even number. Therefore, it is not possible to write a Pythagorean triple with two even numbers and one odd number. All the sides of a Pythagorean triple must have the same parity (i.e., either all even or all odd).