Question

2. If a, b, c are distinct integers that are a Pythagorean triple, why are ka, kb, kc also a Pythagorean triple for any integer k? Written like a 8thgrader

Answer 1

To understand why \(ka\), \(kb\), and \(kc\) form a Pythagorean triple when \(a\), \(b\), and \(c\) are distinct integers that are already a Pythagorean triple, we first need to remember what a Pythagorean triple is.

A Pythagorean triple consists of three numbers \(a\), \(b\), and \(c\) such that:

\[
a^2 + b^2 = c^2
\]

where \(c\) is the longest side of a right triangle.

Now, if we multiply each of these numbers by some integer \(k\), we get new numbers: \(ka\), \(kb\), and \(kc\). We need to check if these new numbers also form a Pythagorean triple:

1. Let's find \( (ka)^2 + (kb)^2 \):

\[
(ka)^2 + (kb)^2 = k^2a^2 + k^2b^2 = k^2(a^2 + b^2)
\]

2. Now let's find \( (kc)^2 \):

\[
(kc)^2 = k^2c^2
\]

3. We know from the original Pythagorean triple that \( a^2 + b^2 = c^2 \). So, we can replace \( a^2 + b^2 \) with \( c^2 \):

\[
k^2(a^2 + b^2) = k^2c^2
\]

4. Thus, we have:

\[
(ka)^2 + (kb)^2 = k^2c^2
\]

This shows that:

\[
(ka)^2 + (kb)^2 = (kc)^2
\]

Since this equation holds true, we can conclude that \(ka\), \(kb\), and \(kc\) also form a Pythagorean triple.

So, when you multiply a Pythagorean triple by the same integer \(k\), the new numbers are still a Pythagorean triple!