Exercise 5.3.8 (There is a Pythagorean triple with $n$ as one of its members)

Question

If $n$ is any integer $\geq 3$, show that there is a Pythagorean triple with $n$ as one of its members

Richard Ganaye · Accepted Answer

Note that $n$ is always member of the Pythagorean triple $(n ,0, n)$, so we show that there is a {\bf positive} Pythagorean triple with $n$ as one of its members.

\begin{proof}

Suppose that $n\geq 3$.
\begin{itemize}
\item If $n$ is odd, then by Problem 7,
$$n^2 = \left(\frac{n^2+1}{2}\right)^2 - \left(\frac{n^2-1}{2}\right)^2,$$
so $\left(n, \frac{n^2-1}{2}, \frac{n^2+1}{2}\right)$ is a (positive) Pythagorean triple with $n$ as one of its members.
\item If $n$ is even, then $n^2 \equiv 0 \pmod 4$, thus
$$n^2 = \left(\left(\frac{n}{2}\right)^2 + 1 \right)^2 -  \left(\left(\frac{n}{2}\right)^2 - 1 \right)^2,$$
so $\left(n, \left(\frac{n}{2}\right)^2 + 1,\left(\frac{n}{2}\right)^2 - 1 \right)$ is a (positive) Pythagorean triple with $n$ as one of its members.
\end{itemize} 
Examples: For $n = 5$, we obtain the triple $(5, 12, 13)$, and for $n = 6$, we obtain $(6, 8, 10)$ (not primitive).
\end{proof}

Exercise 5.3.8 (There is a Pythagorean triple with $n$ as one of its members)

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Exercise 5.3.8 (There is a Pythagorean triple with $n$ as one of its members)

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