Question

question #3 reference q.46918 a 3,4,5 triangle (a triangle with side lengths of 3, 4, and 5) is always a right triangle, and so is a 5,12,13 triangle. are there more of these “pythagorean triples”? what would some of them be?

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with side lengths \(a\), \(b\), and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). Pythagorean triples are positive - integer solutions of this equation.

Step2: Use a formula to generate triples

One way to generate Pythagorean triples is by using the formulas \(a = k(m^{2}-n^{2})\), \(b = 2kmn\), and \(c=k(m^{2}+n^{2})\), where \(m>n>0\) are positive integers, \(m\) and \(n\) are relatively prime and of opposite parity (one is even and one is odd), and \(k\) is a positive integer.
Let \(k = 1\), \(m = 2\), \(n = 1\):
\[

$$\begin{align*} a&=m^{2}-n^{2}=2^{2}-1^{2}=4 - 1=3\\ b&=2mn=2\times2\times1 = 4\\ c&=m^{2}+n^{2}=2^{2}+1^{2}=4 + 1=5 \end{align*}$$

\]
Let \(m = 3\), \(n = 2\):
\[

$$\begin{align*} a&=m^{2}-n^{2}=3^{2}-2^{2}=9 - 4 = 5\\ b&=2mn=2\times3\times2=12\\ c&=m^{2}+n^{2}=3^{2}+2^{2}=9 + 4=13 \end{align*}$$

\]
Let \(m = 4\), \(n = 1\):
\[

$$\begin{align*} a&=m^{2}-n^{2}=4^{2}-1^{2}=16 - 1 = 15\\ b&=2mn=2\times4\times1=8\\ c&=m^{2}+n^{2}=4^{2}+1^{2}=16 + 1=17 \end{align*}$$

\]
Let \(m = 4\), \(n = 3\):
\[

$$\begin{align*} a&=m^{2}-n^{2}=4^{2}-3^{2}=16 - 9 = 7\\ b&=2mn=2\times4\times3=24\\ c&=m^{2}+n^{2}=4^{2}+3^{2}=16 + 9=25 \end{align*}$$

\]

Answer:

Yes, there are infinitely many Pythagorean triples. Some examples are \((8,15,17)\), \((7,24,25)\), \((20,21,29)\)