Is it possible to write a Pythagorean triple in which 2 of the numbers are odd and one is even? Explain your reasoning.
To determine if it is possible to write a Pythagorean triple with two odd numbers and one even number, we need to consider the properties of Pythagorean triples.
A Pythagorean triple consists of three positive integers: a, b, and c. These integers satisfy the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b).
To check if it is possible to have two odd numbers and one even number in a Pythagorean triple, we can consider the squares of these numbers.
Let's assume that the even number is represented by 2n, where n is an integer. Since 2n is even, the square of 2n is divisible by 4.
Now, let's assume that the two odd numbers are represented by 2m + 1 and 2k + 1, where m and k are integers. The squares of these odd numbers are (2m + 1)^2 and (2k + 1)^2.
Expanding these squares, we get:
(2m + 1)^2 = 4m^2 + 4m + 1
(2k + 1)^2 = 4k^2 + 4k + 1
Notice that both of the odd squares have a remainder of 1 when divided by 4. This is because the square of any odd number is always congruent to 1 modulo 4.
Now, according to the Pythagorean theorem, the sum of the squares of the two odd numbers should be equal to the square of the even number. In other words, (4m^2 + 4m + 1) + (4k^2 + 4k + 1) should be divisible by 4.
However, if we simplify this expression, we get:
4m^2 + 4m + 1 + 4k^2 + 4k + 1 = 4(m^2 + m + k^2 + k) + 2
This expression is not divisible by 4 since it has a remainder of 2 when divided by 4. Therefore, it is not possible to write a Pythagorean triple with two odd numbers and one even number.
In conclusion, based on the properties of Pythagorean triples and the behavior of odd and even numbers when squaring them, it is not possible to find a Pythagorean triple where two numbers are odd and one is even.