Is it possible to write a Pythagorean triple in which all of the numbers are even? Explain your reasoning.

To determine if it is possible to write a Pythagorean triple in which all of the numbers are even, we need to understand 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 (according to the Pythagorean theorem). In a Pythagorean triple, one of the numbers has to be odd, and the other two must be even.

Let's assume all three numbers in the Pythagorean triple are even. In this case, we can express the even numbers as 2n, where n is an integer.

Using this notation, the equation becomes (2n)^2 + (2n)^2 = c^2. Simplifying, we get 4n^2 + 4n^2 = c^2. Combining like terms, it becomes 8n^2 = c^2.

Now, we can conclude that c^2 must be divisible by 8 since it equals 8n^2. This means c must also be divisible by √8, or approximately 2.83. However, an even number can only be divisible by another even number. Since 2.83 is not an even number, it cannot be a divisor of c. Therefore, it is not possible to write a Pythagorean triple in which all of the numbers are even.

In summary, because one of the numbers in a Pythagorean triple must be odd, it is not possible to write a Pythagorean triple in which all of the numbers are even.