Is it possible to write a Pythagorean triple in which all of the numbers are odd? Explain your reasoning.
To determine if it is possible to write a Pythagorean triple with all odd numbers, we need to understand what a Pythagorean triple is.
A Pythagorean triple consists of three positive integers (a, b, c) that 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). Mathematically, it can be expressed as:
a^2 + b^2 = c^2
Now, let's consider the properties of odd numbers. An odd number can be represented as 2n + 1, where n is an integer. Squaring an odd number yields:
(2n + 1)^2 = (4n^2 + 4n + 1)
When expanding this expression, we can observe that it will have both even and odd terms, with the odd term being 1. Similarly, if we square any even number, the resulting expression will only consist of even terms.
Now, let's assume that a Pythagorean triple with all odd numbers exists, such that a = 2n + 1, b = 2m + 1, and c = 2k + 1, where n, m, and k are integers.
Substituting these values into the Pythagorean theorem equation, we get:
(2n + 1)^2 + (2m + 1)^2 = (2k + 1)^2
Expanding and simplifying this expression, we have:
4n^2 + 4n + 1 + 4m^2 + 4m + 1 = 4k^2 + 4k + 1
Cancelling out the common factor of 4, we get:
n^2 + n + m^2 + m + 1 = k^2 + k + 1
Rearranging the terms, we have:
(n^2 + n) + (m^2 + m) = (k^2 + k)
Now, let's consider the parity (evenness or oddness) of each term. The sum of any two odd numbers or two even numbers is always even. Similarly, the sum of an odd number and an even number is always odd.
From the equation above, the left-hand side consists of terms (n^2 + n) and (m^2 + m), both of which are even since they involve multiplication of even numbers. This means the left-hand side is even.
On the right-hand side, the term (k^2 + k) must also be even. This can only happen if both k^2 and k are even or both are odd. Now, let's consider these two possibilities:
1. If k^2 and k are both even: This implies that k = 2p for some integer p. Substituting this into (k^2 + k), we have:
(2p^2 + 2p) + (2p) = 4p^2 + 4p
This means the right-hand side is even.
2. If k^2 and k are both odd: This implies that k = 2q + 1 for some integer q. Substituting this into (k^2 + k), we have:
((2q + 1)^2) + (2q + 1) = (4q^2 + 4q + 1) + (2q + 1) = 4q^2 + 6q + 2
In this case, the right-hand side is even.
In both cases, we see that the right-hand side is even, while the left-hand side is odd. This contradiction shows that it is not possible to write a Pythagorean triple in which all of the numbers are odd.
Therefore, it is not possible to have a Pythagorean triple with all odd numbers.