\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
An integer rational point: \((1, 1)\). Notice that \begin {align*} & \phantom {=} (\cos \theta + \sqrt {m} \sin \theta )^2 + (\sin \theta - \sqrt {m} \cos \theta )^2 \\ & = \cos ^2\theta + 2\sqrt {m} \sin \theta \cos \theta + m\sin ^2\theta + \sin ^2\theta -2\sqrt {m}\sin \theta \cos \theta + m\cos ^2\theta \\ & = (m + 1) (\sin ^2\theta + \cos ^2\theta ) \\ & = m + 1. \end {align*}
Consider letting \(x = \cos \theta + \sqrt {m} \sin \theta \), and \(y = \sin \theta - \sqrt {m} \cos \theta \). Let \(m = 1\), and we have \(x = \cos \theta + \sin \theta \) and \(y = \sin \theta - \cos \theta \), with \(x^2 + y^2 = m + 1 = 2\).
Let \(\cos \theta = \frac {3}{5}\), and \(\sin \theta = \frac {4}{5}\). We have \[ (x, y) = \left (\frac {7}{5}, \frac {1}{5}\right ) \] is a non-integer rational point.
An integer \(2\)-rational point: \((1, \sqrt {2})\).
For the non-integer \(2\)-rational point, let \(m = \sqrt {2}\) in the previous question, and we have \[ (\cos \theta + \sqrt {2} \sin \theta )^2 + (\sin \theta - \sqrt {2} \cos \theta )^2 = 2 + 1 = 3. \]
Now, let \(\cos \theta = \frac {3}{5}\) and \(\sin \theta = \frac {4}{5}\). Let \(x = \cos \theta + \sqrt {2} \sin \theta = \frac {3}{5} + \sqrt {2} \cdot \frac {4}{5}\) and \(y = \sin \theta - \sqrt {2} \cos \theta = \frac {4}{5} - \sqrt {2} \cdot \frac {3}{5}\). We must have \(x^2 + y^2 = 3\), and \[ (x, y) = \left (\frac {3}{5} + \sqrt {2} \cdot \frac {4}{5}, \frac {4}{5} - \sqrt {2} \cdot \frac {3}{5}\right ) \] is a non-integer \(2\)-rational point on \(x^2 + y^2 = 3\).
Consider \(x = a \cos \theta + b \sqrt {m} \sin \theta \) and \(y = a \sin \theta - b \sqrt {m} \cos \theta \), we have \begin {align*} x^2 + y^2 & = \left (a \cos \theta + b \sqrt {m} \sin \theta \right )^2 + \left (a \sin \theta - b \sqrt {m} \cos \theta \right )^2 \\ & = a^2 \cos ^2\theta + b^2 m \sin ^2\theta + 2ab\sqrt {m} \sin \theta \cos \theta \\ & \phantom {=} + a^2 \sin ^2\theta + b^2 m \cos ^2\theta - 2ab\sqrt {m} \sin \theta \cos \theta \\ & = (a^2 + b^2 m) \cos ^2\theta + (a^2 + b^2 m) \sin ^2\theta \\ & = (a^2 + b^2 m) (\sin ^2\theta + \cos ^2\theta ) \\ & = a^2 + b^2 m. \end {align*}
We set \(m = 2\), and hence we would like \(a^2 + 2 b^2 = 11\). Consider \(a = 3\) and \(b = 1\), and set \(\cos \theta = \frac {4}{5}\) and \(\sin \theta = \frac {3}{5}\). Hence, \[ x = a \cos \theta + b \sqrt {m} \sin \theta = 3 \cdot \frac {4}{5} + 1 \cdot \sqrt {2} \cdot \frac {3}{5} = \frac {12}{5} + \sqrt {2} \cdot \frac {3}{5}, \] and \[ y = a \sin \theta - b \sqrt {m} \cos \theta = 3 \cdot \frac {3}{5} - 1 \cdot \sqrt {2} \cdot \frac {4}{5} = \frac {9}{5} - \sqrt {2} \cdot \frac {4}{5}, \] and we must have \(x^2 + y^2 = 3^2 + 1^2 \cdot 2 = 11\). Therefore, \[ (x, y) = \left (\frac {12}{5} + \sqrt {2} \cdot \frac {3}{5}, \frac {9}{5} - \sqrt {2} \cdot \frac {4}{5}\right ) \] is a non-integer \(2\)-rational point on the circle \(x^2 + y^2 = 11\).
Consider \(x = a \sec \theta + b \sqrt {m} \tan \theta \) and \(y = a \tan \theta + b \sqrt {m} \sec \theta \), we have \begin {align*} x^2 - y^2 & = \left (a \sec \theta + b \sqrt {m} \tan \theta \right )^2 - \left (a \tan \theta + b \sqrt {m} \sec \theta \right )^2 \\ & = a^2 \sec ^2 \theta + b^2 m \tan ^2\theta + 2ab\sqrt {m} \sec \theta \tan \theta \\ & \phantom {=} - a^2 \tan ^2\theta - b^2 m \sec ^2\theta - 2ab\sqrt {m} \sec \theta \tan \theta \\ & = a^2 (\sec ^2 \theta - \tan ^2 \theta ) - b^2 m (\sec ^2 \theta - \tan ^2 \theta ) \\ & = a^2 - b^2 m. \end {align*}
We set \(m = 2\), and hence we would like \(a^2 - 2 b^2 = 7\). Consider \(a = 3\) and \(b = 1\), and set \(\tan \theta = \frac {3}{4}\) and \(\sec \theta = \frac {5}{4}\). Hence, \[ x = a \sec \theta + b \sqrt {m} \tan \theta = 3 \cdot \frac {5}{4} + 1 \cdot \sqrt {2} \cdot \frac {3}{4} = \frac {15}{4} + \sqrt {2} \cdot \frac {3}{4}, \] and \[ y = a \tan \theta + b \sqrt {m} \sec \theta = 3 \cdot \frac {3}{4} + 1 \cdot \sqrt {2} \cdot \frac {5}{4} = \frac {9}{4} + \sqrt {2} \cdot \frac {5}{4}, \] and we must have \(x^2 - y^2 = 3^2 - 1^2 \cdot 2 = 7\). Therefore, \[ (x, y) = \left (\frac {15}{4} + \sqrt {2} \cdot \frac {3}{4}, \frac {9}{4} + \sqrt {2} \cdot \frac {5}{4}\right ) \] is a non-integer \(2\)-rational point on the hyperbola \(x^2 - y^2 = 7\).