Problems

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Solution: Suppose \(c = a+b\) then \begin{align*} (a+b+c)^3 &-6(a+b+c)(a^2+b^2+c^2) +8(a^3+b^3+c^3) \\ &= (2(a+b))^3-6(2(a+b))(a^2+b^2+(a+b)^2) + 8(a^3+b^3+(a+b)^3) \\ &=16(a+b)^3 - 24(a+b)(a^2+b^2+ab)+8(a^3+b^3) \\ &= 8(a+b)(2(a+b)^2-3(a^2+b^2+ab)+(a^2-ab+b^2)) \\ &= 0 \end{align*} Therefore \(a+b-c\) is a factor. By symmetry \(a-b+c\) and \(-a+b+c\) are also factors. Since our polynomial is degree \(3\) it must be \(K(a+b-c)(b+c-a)(c+a-b)\) for some \(K\). Since the coefficient of \(a^3\) is \(3\), \(K = 3\). so we have: \(3(a+b-c)(b+c-a)(c+a-b)\)

1. We want \(x + a + b = x+1\), \(x^2 + a^2 + b^2 = x^3+\frac52, x^3 + a^3 + b^3 = x^3+ \frac{13}{4}\). \(a+b = 1, a^2 + b^2 = 5/2\) so \(a = \frac32, b = -\frac12\) \begin{align*} 0 &= (x+1)^3 - 3(x+1)(2x^2+5)+2(4x^3+13) \\ &= 3(x +\frac{3}{2}+\frac{1}{2})(x - \frac{3}{2} - \frac{1}{2})(-x + \frac{3}{2} - \frac{1}{2}) \\ &= 3(x+2)(x-2)(1-x) \end{align*} and so the roots are \(x = 1, 2, -2\)
2. Letting \(c = d+e\) we have \begin{align*} (a+b+d+e)^3 &-6(a+b+d+e)(a^2+b^2+d^2+e^2) +8(a^3+b^3+d^3+e^3) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2-2de) +8(a^3+b^3+c^3 - 3cde) \\ &= (a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)+12(a+b+c)de - 24cde \\ &= \underbrace{(a+b+c)^3 -6(a+b+c)(a^2+b^2+c^2)+8(a^3+b^3+c^3)}_{\text{has a factor of }a+b-c} + 12(a+b-c)de \end{align*} Therefore there is a factor of \(a+b-c\) or \(a+b-d-e\). By symmetry we must have the factors: \((a+b-d-e)(a-b-d+e)(a-b+d-e)\) and so the final expression must be: \(K(a+b-d-e)(a-b-d+e)(a-b+d-e)\) The coefficient of \(a^3\) is \(3\), therefore \(K = 3\) We want \(x+a+b+c = x + 6\), \(x^2+a^2+b^2+c^2 = 14\) and \(x^3 + a^3+b^3+c^3 = 36\), ie \(a = 1,b=2,c=3\) would work, so \begin{align*} 0 &= (x+6)^3 - 6(x+6)(x^2+14) +8(x^3+36) \\ &= 3(x+1-2-3)(x-1+2-3)(x-1-2+3) \\ &= 3x(x-4)(x-2) \end{align*} ie the roots are \(x = 0, 2, 4\)

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Solution:

1. \(\,\) Let \(P(x)\) mean the proposition that \(f(x) + (1-x)f(-x) = x^2\) so \begin{align*} P(x): && f(x) + (1-x)f(-x) &= x^2 \\ P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\ \Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\ \Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\ \Rightarrow && f(x)x^2 &= x^3 \\ \Rightarrow && f(x) &= x \end{align*} Notice that \(x + (1-x)(-x) = x^2\) so it does satisfy the functional equation.
2. Let \(K(x) = \frac{x+1}{x-1}\) if \(x \neq 1\) so \begin{align*} && K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\ &&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\ &&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\ &&&= x \end{align*} Let \(Q(x)\) denote the proposition that \(g(x) + xg(K(x)) = x\) so \begin{align*} Q(x): && g(x) + xg(K(x)) &= x \\ Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\ \Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\ \Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\ \Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\ \Rightarrow && g(x) &= \frac{2x}{x^2+1} \end{align*}. And notice that \(\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x\)
3. Consider \(H(x) = \frac{1}{1-x}\) then notice that \(H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}\) and \(H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x\). So So letting \(S(x)\) be the statement that \(h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}\) we have \begin{align*} S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\ S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\ S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\ S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\ \Rightarrow && h(x)& = \frac12 - x \end{align*} and notice that \(\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}\) so it does satisfy the equation.

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Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

   + - ↺
   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

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Solution:

1. Suppose \(X*Y = Y*X\), then \begin{align*} && X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\ && Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\ \Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y}) \end{align*} Therefore, either \(\mathbf{x} = \mathbf{y}\) or \(\lambda = \frac12\). Since we assumed \(X,Y\) were distinct, \(\mathbf{x} \neq \mathbf{y}\) and so \(X*Y\) and \(Y*X\) are distinct unless \(\lambda = \frac12\)
2. Suppose \((X*Y)*Z = X*(Y*Z)\) \begin{align*} &&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\ &&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\ &&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\ &&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\ \Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\ &&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\ &&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x}) \end{align*} Therefore they are distinct unless \(\lambda = 1, 0\) or \(\mathbf{x} = \mathbf{z}\).
3. Claim: \((X*Y)*Z = (X*Z)*(Y*Z)\) Proof: \begin{align*} && (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ && (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z} \end{align*} Claim: \(X*(Y*Z) = (X*Y)*(X*Z)\) Proof: \begin{align*} X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ (X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \end{align*}
4. \(P_1 = X*Y\) divides the line segment into the ratio \(\lambda:(1-\lambda)\). \(P_n\) divides the line segment \(P_{n-1}Y\) into the ratio \(\lambda:(1-\lambda)\), therefore it divides the line segment \(XY\) in the ratio \(\lambda^n : 1- \lambda^n\) Alternatively, \begin{align*} P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\ P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\ &= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y} \end{align*} Suppose \(P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}\) then \begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\ &= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\ & = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}

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Solution:

1. \(\,\) \begin{align*} && |z-w|^2 &= (z-w)(z^*-w^*) \\ &&&= zz^* - wz^*-zw^* + ww^* \\ &&&= |z|^2+|w|^2 - E + 2|zw| \\ &&&= (|z|+|w|)^2 - E \\ \Rightarrow && E &= (|z|+|w|)^2 - |z-w|^2 &\in \mathbb{R} \end{align*} and by the triangle inequality \(|z|+|w| \geq |z-w|\), so \(E \geq 0\)
2. \(\,\) \begin{align*} && |1-zw^*|^2 &= (1-zw^*)(1-z^*w) \\ &&&= 1 - zw^*-z^*w + |zw|^2 \\ &&&= 1 - E + 2|zw| + |zw|^2 \\ &&&= (1+|zw|)^2 - E \end{align*} \begin{align*} && \frac{|z-w|^2}{|1-zw^*|^2} &= \frac{(|z|+|w|)^2-E}{(1+|zw|)^2-E} \\ \Leftrightarrow && (1+|zw|^2)|z-w|^2 -E|z-w|^2 &= (|z|+|w|)^2|1-zw^*|^2-E|1-zw^*|^2\\ \Leftrightarrow && (1+|zw|^2)|z-w|^2-(|z|+|w|)^2|1-zw^*|^2 &= E(|z-w|^2-|1-zw^*|^2)\\ &&&= E(|z|^2-zw^*-z^*w+|w|^2-1+zw^*+z^*w-|z|^2|w|^2) \\ &&&= E(|z|^2+|w|^2-1-|z|^2|w|^2) \\ &&&= -E(1-|z|^2)(1-|w|^2) \\ &&&\leq 0 \\ \Leftrightarrow&& (1+|zw|^2)|z-w|^2& \leq (|z|+|w|)^2|1-zw^*|^2\\ \Leftrightarrow&& \frac{|z-w|^2}{|1-zw^*|^2} &\leq \frac{(|z|+|w|)^2}{(1+|zw|)^2}\\ \Leftrightarrow && \frac{|z-w|}{|1-zw^*|} &\leq \frac{(|z|+|w|)}{(1+|zw|)}\\ \end{align*} Yes, this inequality holds if \(|z|, |w|\) are the same side of \(1\) and is reversed otherwise.

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Solution:

1. Suppose \(t_n = x\) for all \(n\), then we must have \begin{align*} && x &= p x + q x \\ \Leftrightarrow && 1 &= p+q \end{align*} and this clearly works for any value of \(x\).
2. Suppose \(t_{2n} = x, t_{2n+1} = y\) for all \(n\), then \begin{align*} && x &= py + q x \\ && y &= px + q y \\ \Rightarrow && 0 &= py + (q-1) x \\ && 0 &= px + (q-1) y \\ \Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\ \Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\ \Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\ \end{align*}
3. Suppose \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) , so \begin{align*} && x &= pz + qy \\ && y & = px + qz \\ && z &= py + qx \\ \\ && z &= p(px+qz) + q(pz + qy) \\ &&&= p^2x + 2pqz + q^2 y \\ &&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\ &&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\ &&&= (p^3+q^3+2pq)z + pq(py+qx) \\ &&&= (p^3 + q^3 + 2pq)z + pq z \\ &&&= (p^3 + q^3 + 3pq)z \\ \Rightarrow && 0 &= p^3 + q^3 + 3pq- 1 \\ &&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\ &&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2) \end{align*} Therefore \(p+q = 1\) or \((p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1\). If \(p+q = 1\), then \(z = py + (1-p)x\) and \(x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y \Rightarrow x = y \Rightarrow x= y = z\). If \(p = q = -1\) then adding all the equations we get \(x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0\)

Note that what is actually going on here is that solutions must be of the form \(t_n = \lambda^n\) so the only way to be constant is for \(\lambda = 1\) to be a root, the only way for it to be \(2\)-periodic is for \(\lambda = -1\) to be a root, and the only way for it to be \(3\)-periodic is for \(\lambda = 1, \omega, \omega^2\) to be the roots (although we see this via the classic \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)\) which is because of the real constraint in the question.

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Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

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Solution:

\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

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Solution: \begin{align*} && \beta - \alpha &> q \\ \Rightarrow &&(\beta - \alpha)^2 &> q^2 \\ \Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\ \Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\ \Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0 \end{align*} \begin{align*} && u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\ &&&= u_n^2 + u_{n+1}^2-q^2 \\ &&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\ &&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\ &&&= u_{n+1}(u_{n-1}+u_{n+1}) \\ \\ && u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\ &&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}} \end{align*} Therefore if \(u_2 -pu_1 + u_0 = 0\) it is always zero, ie if \begin{align*} && u_2 &= p\beta - \alpha \\ && u_2 &= \frac{\beta^2-q^2}{\alpha} \\ \Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\ \Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} \end{align*} If \(\beta > \alpha + q\) we must have that \(p > 2\), and so \(u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0\), therefore the sequence is strictly increasing. If \(\beta = \alpha + q\) the sequence follows \(u_{n+1} - 2u_n + u_{n-1} =0\) and so \(u_{n+1}-u_n = u_n - u_{n-1}\) for all \(n\) (which is still increasing - it's an arithmetic progression with common difference \(\beta - \alpha\)).

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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 \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)

Note: we can also find the additional point in the last part by considering lines through \((3, \sqrt2)\), for example \(y = -\frac32x + \sqrt2 + \frac92\) would give the same point.

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Solution:

1. \(\,\) \begin{align*} && 1 &= \frac{a}{x} + \frac{b}{y} \\ \frac{\d}{\d x}: && 0 &= -\frac{a}{x^2} - \frac{b}{y^2} \frac{\d y}{\d x} \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{ay^2}{bx^2} \\ \\ (p,q): && -\frac{aq^2}{bp^2} &= -\frac{a}{b} \\ \Rightarrow && p^2 &= q^2 \\ \Rightarrow && p &= \pm q \\ \\ \Rightarrow && ap \pm b p &= 1 \\ \Rightarrow && (a\pm b)p &= 1 \\ \Rightarrow && \frac{a}{p} \pm \frac{b}{p} &= 1 \\ \Rightarrow && (a \pm b)\frac{1}{p} &= 1 \\ \Rightarrow && (a \pm b)^2 &= 1 \end{align*}
2. \(\,\) \begin{align*} && 1 &= \frac{a}{x} - \frac{b}{y} \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{ay^2}{bx^2} \\ \Rightarrow && \frac{aq^2}{bp^2} &= \frac{b}{a} \\ \Rightarrow && aq &= \pm bp \\ \Rightarrow && 1 &= \frac{a}{p} - \frac{b}{q} \\ &&&= \frac{aq-bp}{pq} \\ \Rightarrow && aq &= -bp \\ \Rightarrow && 1 &= \frac{2aq}{pq} \\ \Rightarrow && p &= 2a \\ \Rightarrow && q &= -2b \\ \Rightarrow && 1 &= 2a^2-2b^2 \\ \Rightarrow && \frac12 &= a^2-b^2 \end{align*}

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Solution: \begin{align*} \int_0^1 \frac{x \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x+1-1) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( e^x -\frac{\e^x}{1+x} \right )\, \d x \\ &= \e-1-E \end{align*} \begin{align*} \int_0^1 \frac{x^2 \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x^2+x-x) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( xe^x -\frac{x\e^x}{1+x} \right )\, \d x \\ &= \left [xe^{x} \right]_0^1 - \int_0^1 e^x \, \d x -(\e-1-E) \\ &= \e-(\e-1)-(\e -1 -E) \\ &= 2-\e + E \end{align*}

1. Since \(\displaystyle u = \frac{1-x}{1+x},\frac{\d u}{\d x} = \frac{-(1+x)-(1-x)}{(1+x)^2}\), \begin{align*} && \int_0^1 \frac{\e^{\frac{1-x}{1+x}}}{1+x}\, \d x &= \int_{u=1}^{u=0} \frac{e^u}{1+x} \cdot \frac{(1+x)^2}{-2} \d u \\ &&&= \int_0^1 \frac{(1+x) e^u}{2} \d u \\ &&&= \int_0^1 \frac{\left ( 1 + \frac{1-u}{1+u} \right) e^u}{2}\, \d u \\ &&&= \frac12 \int_0^1 \left (e^u + \frac{e^{u}}{1+u} - \frac{ue^u}{1+u} \right) \, \d u \\ &&&= \frac12 \left( \e-1 + E - (\e - 1 - E) \right) \\ &&&= E \end{align*}
2. Since \(\displaystyle u = x^2-1, \d u = 2x \d x\)\begin{align*} \int_1^{\sqrt2} \frac {\e^{x^2}}x \, \d x &= \int_{u=0}^{u=1} \frac{e^{u+1}}{x} \frac{1}{2x} \d u \\ &= \int_0^1 \frac{e^{u+1}}{2(u+1)} \d u \\ &= \frac{\e}{2} E \\ &= \frac{E\e}{2} \end{align*}

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Solution:

1. 

   + - ↺
   Clearly the only solution is \(x = 1\)
2. 

   + - ↺
   There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

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Solution: \begin{array}{c|c} n & n^3 \\ \hline 1 & 1 \\ 2 & 8 \\ 3 & 27 \\ 4 & 64 \\ 5 & 125 \\ 6 & 216 \\ 7 & 343 \\ 8 & 512 \\ 9 & 729 \\ 10 & 1000 \\ \end{array}

1. \(\,\) \begin{align*} && x^3 + y^3 &= kz^3 \\ \Rightarrow &&k(x^2-xy+y^2)&=kz^3 \\ \Rightarrow && z^3 &= (x+y)^2-3xy \\ &&&= k^2-3x(k-x) \\ &&&= k^2-3xk+3x^2 \\ \\ \Rightarrow && \frac{4z^3-k^2}{3} &= \frac{4(k^2-3xk+3x^2)-k^2}{3} \\ &&&= \frac{3k^2-12xk+12x^2}{3} \\ &&&= k^2-4xk+4x^2 \\ &&&= (k-2x)^2 \end{align*} Therefore \(\frac{4z^3-k^2}{3}\) is a perfect square and so \(4z^3 \geq k^2 \Rightarrow z^3 \geq \frac14k^2\). Clearly \(kz^3 < x^3+3x^2y+3xy^2+y^3 = k^3 \Rightarrow z^3 < k^2\), therefore \(\frac14 k^2 \leq z^3 < k^2\) Therefore if \(k = 20\), \(100 \leq z^3 < 400 \Rightarrow z \in \{ 5, 6,7\}\). Mod \(3\) it is clear that \(4z^3-k^2\) is not divisible by \(3\) for \(z = 5,6\) therefore \(z = 7\) \begin{align*} && 343 &= 3x^2-60x+400 \\ \Rightarrow && 0 &= 3x^2-60x+57 \\ \Rightarrow && 0 &= x^2-20x+19 \\ \Rightarrow && x &= 1,19 \end{align*} Therefore a solution is \(1^3 + 19^3 = 20 \cdot 7^3\)
2. When \(x+y = z^2\) we must have \begin{align*} && x^3 + y^3 &= kz^3 \\ \Rightarrow &&(x^2-xy+y^2)&=kz \\ \Rightarrow && kz &= (x+y)^2-3xy \\ &&&= z^4-3x(z^2-x)\\ &&&= z^4-3xz^2+3x^2 \\ \Rightarrow && 0 &= 3x^2-3z^2x+z^4-kz \\ \\ \Rightarrow && 0 &\leq \Delta = 9z^4-12(z^4-kz) \\ &&&=12kz-3z^4 \\ \Rightarrow && z^3 &\leq 4k \end{align*} If \(k = 19\) this means \(z \leq 4\) \begin{array}{c|c|c|c} z & 19z^3 & x & y \\ \hline 1 & 19 & - & - \\ 2 & 152 & 3 & 5 \\ 3 & 513 & 1 & 8 \end{array} So two solutions are \(1^3+8^3 = 19 \cdot 3^3\) and \(3^3+5^3=19 \cdot 2^3\)