Problems

Solution: \begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}

1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}

Solution: \begin{align*} I_n - I_{n+1} &= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u - \int_0^\infty \frac 1 {(1+u^2)^{n+1}}\, \d u \\ &= \int_0^\infty \l \frac 1 {(1+u^2)^n}- \frac 1 {(1+u^2)^{n+1}} \r\, \d u \\ &= \int_0^\infty \frac {u^2} {(1+u^2)^{n+1}} \, \d u \\ &= \left [ u \frac{u}{(1+u^2)^{n+1}} \right]_0^{\infty} - \frac{-1}{2n}\int_0^{\infty} \frac{1}{(1+u^2)^n} \d u \tag{\(IBP: u = u, v' = \frac{u}{(1+u^2)^{n+1}}\)}\\ &= \frac{1}{2n} I_n \end{align*} \(\displaystyle I_1 = \int_0^{\infty} \frac{1}{1+u^2} \d u = \left [ \tan^{-1} u \right]_0^\infty = \frac{\pi}{2}\) as expected. We also have, \(I_{n+1} = \frac{2n(2n-1)}{2n \cdot 2n} I_n \), by rearranging the recurrence relation. Therefore, when we multiply out the top we will have \(2n!\) and the bottom we will have two factors of \(n!\) and two factors of \(2^n\) combined with the original \(\frac{\pi}{2}\) we get \[ I_{n+1} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2} \] \begin{align*} J = \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{u = \infty}^{u = 0} f((u^{-1}-u)^2)(-u^{-2} )\d u \tag{\(u = x^{-1}, \d u = -x^{-2} \d x\)} \\ &= \int^{u = \infty}_{u = 0} f((u^{-1}-u)^2)u^{-2} \d u \\ &= \int^{\infty}_{0} u^{-2}f((u-u^{-1})^2) \d u \\ \end{align*} Therefore adding the two forms for \(J\) we have \begin{align*} 2 J &= \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x + \int_0^\infty x^{-2} f\big( (x- x^{-1})^2\big ) \, \d x \\ &= \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x \end{align*} And letting \(u = x - x^{-1}\), we have \(\d u = (1 + x^{-2}) \d x\), and \(u\) runs from \(-\infty\) to \(\infty\) so we have: \begin{align*} \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{-\infty}^\infty f(u^2) \, \d u \\ &=2 \int_{0}^\infty f(u^2) \, \d u \end{align*} Since both of these are \(2J\) we have the result we are after. Finally, \begin{align*} \int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x &= \int_0^{\infty} \frac{x^{2n-2}}{x^{2n}(x^2-1+x^{-2})^n} \d x \\ &= \int_0^{\infty} \frac{x^{-2}}{((x-x^{-1})^2+1)^n} \d x \\ &= \int_0^{\infty} \frac{1}{(x^2+1)^n} \d x \tag{Where \(f(x) = (1+x^2)^{-n}\) in \(J\) integral} \\ &= I_n = \frac{(2n-2)! \pi}{2^{2n-1} ((n-1)!)^2} \end{align*}

Solution:

1. If \(m = 1000\), then \(n \geq m \Rightarrow n^2 \geq 1000n \Rightarrow (1000n) \leq (n^2)\)
2. This is false. Let \(s_i = 1,2,1,2,\cdots\) and \(t_i = 2,1,2,1,\cdots\).
3. Suppose that for \(n \geq m_1, s_n \le t_n\) and for \(n \geq m_2, s_t \le u_n\), then for \(n \geq m = \max(m_1, m_2), s_n \leq t_n \leq u_n \Rightarrow s_n \leq u_n \Rightarrow (s_n) \leq (u_n)\)
4. Let \(m = 6\), then if \(n \geq m, 2^n \geq 1 + n + \frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n + 1 = n^2 + n + 2 \geq n^2\), so \((2^n) \geq (n^2)\)

Solution:

1. Let \(z \in \mathbb{R}\) and let \(z \to \pm \infty\) then \(z^3 + az^2 + bz + c\) changes sign, therefore somewhere it must have a real root.
2. \begin{align*} &&z^3 + az^2 + bz + c &= (z-z_1)(z-z_2)(z-z_3) \\ && &= z^3 - (z_1+z_2+z_3)z^2 + (z_1z_2 + z_2z_3+z_3z_1)z - (z_1z_2z_3) \\ \\ \Rightarrow && S_1 &= z_1+z_2+z_3 \\ &&&= -a \\ \\ \Rightarrow && S_2 &= z_1^2+z_2^2+z_3^2 \\ &&&= (z_1+z_2+z_3)^2 - 2(z_1z_2 + z_2z_3+z_3z_1) \\ &&&= a^2 - 2b \\ \Rightarrow && a &= -S_1 \\ && b &= \frac12 \l S_1^2 - S_2\r \\ \\ && 0 &= z_i^3 + az_i^2+bz_i+c \\ \Rightarrow && 0 &= S_3 + aS_2+bS_1+3c \\ &&&= S_3 -S_1S_2 + \frac12 \l S_1^2 - S_2\r S_1 + 3c \\ \Rightarrow && 0 &= 2S_3 - 3S_1S_2 + S_1^3 + 6c \end{align*}
3. Let \(z_k= r_ke^{i \theta_k}\), then we have \(\textrm{Im}(S_k) = 0\) and so the polynomial with roots \(z_k\) has real coefficients, and therefore at least one root is real. This root will have \(\theta_k = 0\). Moreover, since if \(w\) is a root of a real polynomial \(\overbar{w}\) is also a root, and therefore if \(\theta_1 = 0\), we must have that \(z_2\) and \(z_3\) are complex conjugate, ie \(\theta_2 = - \theta_3\)

Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

Solution:

1. \(\,\) \begin{align*} && 3 &= 2^2 - 1^2 \\ && 5 &= 3^2 - 2^2 \\ && 8 &= 3^2 - 1^2 \\ && 16 &= 5^2 - 3^2 \end{align*}
2. Suppose \(n = 2k+1\), then \(n = (k+1)^2 - k^2\)
3. Suppose \(n = 4k\) then \(n = (2k+1)^2 - (2k-1)^2\)
4. All squares leave a remainder of \(0\) or \(1\) on division by \(4\). Therefore the difference can leave a remainder of \(0\), \(1\), \(-1 \equiv 3\), none of which are \(2\).
5. Suppose \(n = pq = a^2 - b^2\) with \(a > b\) ie \((a-b)(a+b) = pq\). Since \(p\) is prime, \(p \mid (a-b)\) or \(p \mid (a+b)\). Similarly for \(q\). Suppose also (wlog) that \(p > q\) Since the factors of \(pq\) are \(1, p, q, pq\) then \(a-b = 1, p\) (which are two possibilities) and \(a+b = pq, q\), ie \(a = \frac{1+pq}{2}, \frac{p+q}{2}\) and \(b = \frac{pq-1}{2}, \frac{p-q}{2}\) \begin{align*} && pq &= \left ( \frac{1+pq}{2} \right)^2- \left ( \frac{1-pq}{2} \right)^2 \\ &&&= \left ( \frac{p+q}{2} \right)^2- \left ( \frac{p-q}{2} \right)^2 \\ \end{align*} Where everything is an integer since \(p\) and \(q\) are odd. If we have \(p > 2\) and \(q = 2\) then \(p\) is odd and the number has the form \(4k+2\) which cannot be expressed as the difference of two squares.
6. \(675 = 3^3 \cdot 5^2\), each factor pair of \(675\) will lead to a different solution of \(675 = a^2-b^2\), since we will have an equation \(a-b = X, a+b = Y\) where \(X, Y\) are both odd. Therefore there are as many solution as (half) the number of factors, ie \(4 \times 3 = 12\)

Solution:

1. Note that \(f(x) = (x+2a)^3 - 27a^2x\) so \(f'(x) = 3(x+2a)^2-27a^2 = 3x^2+12ax-15a^2 = (x-a)(3x+15a)\) so the turning points are at \(x = a, x = -5a\). But \(f(a) = 0\), so the curve just touches the \(x\)-axis. Note also that \(f(0) = 8a^3 \geq 0\) so:
   

   + - ↺
2. \(\,\) \begin{align*} && 0 &\leq (x+2y)^3 - 27y^2 x \\ &&& \leq 3^3 - 27 y^2 x \\ &&&= 27(1-y^2x) \\ \Rightarrow && xy^2 &\leq1 \end{align*} with equality when \(x = y = 1\)
3. Notice that \begin{align*} && 0 &\leq (p+q+r)^3 - 27\left (\frac{q+r}{2}\right)^2 p \\ \Rightarrow && (p+q+r)^3 &\geq 27\left (\frac{q+r}{2}\right)^2 p \\ &&&\underbrace{\geq}_{AM-GM} 27\left (\sqrt{qr}\right)^2 p \\ &&&= 27pqr \end{align*} Equality can only hold if \(p = \frac{q+r}{2}\), but by symmetry we must also have \(q = \frac{r+p}{2}, r = \frac{p+q}{2}\) ie \(p = q = r\). And indeed equality does hold in this case.

Solution: \begin{align*} && \mathbb{P}(X > 4) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n} \\ && \mathbb{P}(X > 3) &= 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \\ \Rightarrow && \mathbb{P}(X =4) &= \mathbb{P}(X > 3) - \mathbb{P}(X > 4) \\ &&&= \frac{(n-1)(n-2)}{n^2} \left (1 - \frac{n-3}{n} \right) \\ &&&= \frac{3(n-1)(n-2)}{n^3} \end{align*}

1. Notice that \begin{align*} && \mathbb{P}(X > r) &= \frac{n-1}{n} \cdots \frac{n-r+1}{n} \\ \Rightarrow && \mathbb{P}(X = r) &= \frac{n-1}{n} \cdots \frac{n-r+2}{n} \left (1 - \frac{n-r+1}{n} \right) \\ &&&= \frac{(n-1)\cdots(n-r+2)(r-1)}{n^{r-1}} \\ &&&= \left (1 - \frac{1}n \right)\left (1 - \frac{2}{n} \right) \cdots \left (1 - \frac{r-2}{n} \right) \frac{r-1}{n} \\ \Rightarrow && 1 &= \sum \mathbb{P}(X = r) \\ &&&= \sum_{r=2}^{n+1} \mathbb{P}(X = r) \\ &&&= \frac 1n + \left(1-\frac1n\right) \frac 2 n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac3 n + \cdots \end{align*}
2. \(\,\) \begin{align*} \mathbb{E}(X) &= \sum_{r=2}^{n+1} r\cdot\mathbb{P}(X = r) \\ &= \frac 2n + \left(1-\frac1n\right) \frac {2\cdot3} n + \left(1-\frac1n\right)\left(1-\frac2n\right) \frac{3\cdot4} n + \cdots \end{align*}
3. \(\displaystyle \mathbb{P}(X \geq k) = \frac{n-1}{n} \cdots \frac{n-r+2}{n}\)
4. \(\,\) \begin{align*} && \mathbb{E}(Y) &= \sum_{r=1}^N r \cdot \mathbb{P}(Y = r) \\ &&&= \sum_{r=1}^N \sum_{j=1}^r \mathbb{P}(Y = r) \\ &&&= \sum_{j=1}^N \sum_{r=j}^N \mathbb{P}(Y=r) \\ &&&= \sum_{j=1}^N \mathbb{P}(Y \geq j) \end{align*} Let \(P_k = \left(1-\frac1n\right)\left(1-\frac2n\right) \cdots \left(1-\frac1n\right)\left(1-\frac{k}n\right) \) \begin{align*} && \mathbb{E}(X) &= P_1 \frac{1 \cdot 2 }{n} + P_2 \cdot \frac{2 \cdot 3}{n} + \cdots + P_k \cdot \frac{k(k+1)}{n} + \cdots \\ && &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + \sum_{k=1}^{n} \frac{k}{n}P_k \\ && \text{Using the identity } & \frac{k}{n}P_k = \frac{k}{n} \prod_{i=1}^{k-1} \left(1-\frac{i}{n}\right) = P_k - P_{k+1}: \\ && \sum_{k=1}^{n} \frac{k}{n}P_k &= (P_1 - P_2) + (P_2 - P_3) + \cdots + (P_n - P_{n+1}) \\ && &= P_1 - P_{n+1} = 1 - 0 = 1 \\ \\ \Rightarrow && \mathbb{E}(X) &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ && &= \mathbb{P}(X \geq 1) + \mathbb{P}(X \geq 2) + \mathbb{P}(X \geq 3) + \cdots \\ && &= 1 + P_1 + P_2 + P_3 + \cdots \\ && &= 1 + \sum_{k=1}^{n} P_k \\ \\ \Rightarrow && 1 + \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k + 1 \\ \Rightarrow && \sum_{k=1}^{n} P_k &= \sum_{k=1}^{n} \frac{k^2}{n}P_k \\ \Rightarrow && 0 &= \sum_{k=1}^{n} P_k \left( 1 - \frac{k^2}{n} \right) \end{align*}

Solution:

1. \((x,y) = (1,0)\) we have Suppose \(p^2-2q^2 = 1\), then \begin{align*} && (3p+4q)^2-2\cdot(2p+3q)^2 &= 9p^2+24pq + 16q^2 - 2\cdot(4p^2+12pq+9q^2) \\ &&&= p^2(9-8) + pq(24-24) + q^2(16-18) \\ &&&= p^2 - 2q^2 = 1 \end{align*} So we have: \begin{array}{c|c} x & y \\ \hline 1 & 0 \\ 3 & 2 \\ 17 & 12 \\ \end{array}
2. Suppose \(x = 2m+1\) and \(y = 2n\) then \begin{align*} && 1 & = x^2 - 2y^2 \\ &&&= (2m+1)^2 - 2(2n)^2 \\ &&&= 4m^2 + 4m + 1 - 8n^2 \\ \Leftrightarrow && n^2 &= \frac{m(m+1)}{2} \end{align*}
3. Suppose \(b^3 = c^4 - a^2 =(c^2-a)(c^2+a)\), since \(b\) is prime and \(c^2 + a > c^2-a\) we must have: \begin{align*} && p = c^2-a && p^2 =c^2 +a \\ \Rightarrow && c^2 = \frac{p+p^2}{2} && a = \frac{p^2-p}2\\ && 1 = c^2-a && p^3 = c^2+a \\ \Rightarrow && c^2 = \frac{p^3+1}{2} && a = \frac{p^3-1}{2} \end{align*} Note that \(c^2 = \frac{p(p+1)}{2}\) is reminicent of our first equation, so suppose \(n = c = 6\) and \(p = m = 8\) then \(6^4 = 8^3 + 28^2\)

Solution:

1. Suppose \(q^nN = p^n\) then since \((p,q) =1\) we must have \(p \mid N\), and then by dividing both \(p^n\) and \(N\) by \(p\) we can repeat this process \(n\) times to find that \(N = kp^n\) and in particular \(q = 1\). Suppose \(\sqrt[n]{N} = \frac{p}q\) for \(p,q\) coprime positive integers (ie it is not irrational), then \(q^nN = p^n\) and so \(q = 1\) and in fact \(\sqrt[n]{N}\) is an integer so \(N\).
2. Suppose \((a,b) = 1, (c,d) = 1\) and \(a^ad^b = b^ac^b\), then since \((a,b) = 1\) we must have \((b^a, a) = 1\) so \(b^a \mid d^b\). Similarly since \((c,d) = 1\) we must have \((d^b, c) = 1\) so \(d^b \mid b^a\). Therefore \(d^b = b^a\). Suppose \(p \mid d\) then \(p \mid d^b = b^a \Rightarrow p \mid b\). Suppose \(\nu_p(d) = m, \nu_p(b) = n\) we must have \(bm = \nu_p(d^b) = \nu_p(b^a) = an\), ie \(b = \frac{an}{m}\). Note that \(p^n \mid b \Rightarrow p^n \mid n \frac{a}{m} \Rightarrow p^n \mid n \Rightarrow p^n \leq n\). Since \((p,a) = 1\).. But since \(p^n > n\) if \(p \geq 2\) we must have that \(b = 1\). Therefore suppose \(r = \frac{a}{b}\) with \((a,b) = 1\) an \(r^r = \frac{c}{d}\) we must ahve \(a^ac^b = b^ad^b\) and so \(b = 1\) implying \(r\) is an integer.

Solution: \begin{align*} \text{COM}: && \sum_{i=1}^n im \cdot \frac{u}{i} &= \left ( km + \sum_{i=1}^n im \right) v \\ \Rightarrow && nu &= \left ( k + \frac{n(n+1)}{2} \right) v \\ \Rightarrow && v &= \frac{2nu}{2k + n(n+1)} \end{align*} If \(2k = N(N+1)\), there will be no more collisions when \(v_n > \frac{u}{n+1}\), ie \begin{align*} && \frac{u}{n+1} &<\frac{2nu}{2k + n(n+1)} \\ \Leftrightarrow && N(N+1) + n(n+1) &< 2n(n+1) \\ \Leftrightarrow && N(N+1) &< n(n+1) \\ \end{align*} Therefore \(n = N+1\) and there will be \(N+1\) collisions. The loss of kinetic energy is: \begin{align*} && \text{initial k.e.} &= \sum_{k=1}^{N+1} \frac12 im \cdot \frac{u^2}{i^2} \\ &&&= \frac12 m u^2 \left ( \sum_{k=1}^{N+1} \frac{1}{i}\right) \\ && \text{final k.e.} &= \frac12 \left ( k + \frac{(N+1)(N+2)}{2}\right)m \left ( \frac{2(N+1)u}{N(N+1)+(N+1)(N+2)} \right)^2 \\ &&&= \frac12 m u^2 \frac{2(N+1)^2}{(N+1)(2N+2)} \\ &&&= \frac12 mu^2 \\ \Rightarrow && \Delta \text{ k.e.} &= \frac12 m u^2 \left ( \sum_{k=2}^{N+1} \frac{1}{i}\right) \end{align*}

Solution:

1. • \(\bf (a)\) \((1, \sqrt2)\) is an integer \(2\)-rational point. \((\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)\) is a non-integer \(2\)-rational point.
   • [\bf(b)] First notice that \((\sqrt2)^2 +3^2 = 11\) so then consider \((1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)\) will work as \(\pi/4\) degree rotation.
   • [\bf(c)] First notice \(3^2-(\sqrt2)^2 = 2\). Notice that \((k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m\). Taking \(k= 3\) we have \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)