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?

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Accepted Answer

# 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 $(8,15,17)$, $(7,24,25)$, $(20,21,29)$

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Explanation:

Step1: Recall Pythagorean theorem

Step2: Use a formula to generate triples

Answer: