<s>[INST] Number Theory with Pythagorean Triples For $n\ge3$ a given integer, find a Pythagorean Triple having n as one of its members.
Hint: For n an odd integer, consider the triple $$\left(n, \frac 12\left(n^2-1\right), \frac 12(n^2+1)\right);$$ For n even, consider the triple $$\left(n, \left(\frac{n^2}{4}\right)-1, \left(\frac{n^2}{4}\right)+1 \right)$$
I have been trying to solve this problem by letting $n=x-y$ or $n=2xy$ but have been unable to use the hints properly.  Do I just add the squares of the first 2 and see if it equals the square of the last?  I am just missing the concept here.  Any suggestions would be appreciated.
$$n^2+\left(\frac{n^2-1}{2}\right)^2=n^2+\frac{n^4-2n^2+1}{4}=\frac{n^4+2n^2+1}{4}$$
$$\left( \frac{n^2+1}{2}\right)^2=\frac{n^4+2n+1}{4}$$
Is this all I have to do for odd?
 [/INST] $\\ \textbf{Matching sides using Euclid's F(m,n)}: $ Solving for $n$, any values of $m$ that yield integers provide the $F(m,n)$ to identify a triple. Examples follow the solutions
$$\mathbf{A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad  \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil}$$
The lower limit ensures $m^2>A$ and the upper limit ensures $m-n\ge 1$.
$$\text{Example:}\qquad  A=21\Rightarrow \lceil\sqrt{21+1}=5 \le m \le \biggl\lceil\frac{21}{2}\biggr\rceil =11\quad and \quad m\in\{5,11\}\Rightarrow n\in\{2,10\} $$
$F(5,2)=(21,20,29)\qquad\qquad F(11,10)=(21,220,221)\\ $
$$\mathbf{B=2mn\Rightarrow n=\frac{B}{2m}\qquad\qquad \bigl\lceil\sqrt{B}\bigr\rceil\le m \le \frac{B}{2}}$$
The lower limit ensures $m>n$ and the upper limit ensures $m-n\ge 1$.
$$\text{Example:}\qquad  B=44\Rightarrow \lceil\sqrt{88}\rceil =10 \le m \le \frac{44}{2}=22\quad and \quad m\in\{11,22\}\Rightarrow n\in\{2,1\}$$
$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)\\$
$$\mathbf{C=m^2+n^2\Rightarrow n=\sqrt{C-m^2}\qquad\qquad \biggl\lceil\sqrt{\frac{C}{2}}\biggr\rceil \le m < \sqrt{C}}$$
$$\text{Example:}\qquad  C=1105\Rightarrow \biggl\lfloor\sqrt{\frac{1105}{2}}\biggr\rfloor=23 \le m < \lfloor\sqrt{1105}=33\quad and \quad m\in\{24,31,32,33\}\Rightarrow n\in\{23,12,9,4\}\\$$
$F(24,23)=(47,1104,1105)\quad F(31,12)=(817,744,1105)\\ $
$F(32,9)=(943,576,1105)\quad F(33,4)=(1073,264,1105)\\$
</s>