Problems

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Solution:

1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

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Solution:

1. Let \(x = \frac{pz+q}{z+1}\) then \begin{align*} && 0 &= x^3-3pqx+pq(p+q) \\ &&&= \left ( \frac{pz+q}{z+1} \right)^3 - 3pq \left ( \frac{pz+q}{z+1} \right) + pq(p+q) \\ &&&= \frac{(pz+q)^3-3pq(pz+q)(z+1)^2+pq(p+q)(z+1)^3}{(z+1)^3} \\ &&&= \frac{1}{(z+1)^3} \Big ((p^3+pq(p+q)-3p^2q)z^3 + (3p^2q-6p^2q+3pq^2+3p^2q+3pq^2)z^2 + \\ &&&\qquad \qquad\quad\quad +(3pq^2-3p^2q-6pq^2+3p^2q+3qp^2)z+(q^3-3pq^2+p^2q+pq^2) \Big ) \\ &&&= \frac{(p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2)}{(z+1)^3} \\ \Rightarrow && 0 &= (p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2) \\ &&&= p(p-q)^2z^3 + q(p-q)^2 \\ \Rightarrow && 0 &= pz^3 + q \end{align*}
2. We would like to find \(pq = c\) and \(pq(p+q) = d\), so \(p\) and \(q\) are roots of the quadratic \(x^2-\frac{d}{c}x + c = 0\), which has distinct real roots if \(\Delta = \frac{d^2}{c^2}-4c > 0 \Rightarrow d^2>4c^3\)
3. Note that \(c = -2, d = -2\) so \begin{align*} && 0 &= x^3+6x-2 \\ \text{consider} && 0 &= X^2-X-2 \\ && &= (X+1)(X-2) \\ \Rightarrow && p = -1, &q = 2\\ \Rightarrow && 0 &= x^3-3\cdot 2 \cdot(-1) x + 2\cdot(-1) \cdot(-2+1) \\ \Rightarrow && 0 &= -z^3+2 \\ \Rightarrow && z &= \sqrt[3]{2} \\ \Rightarrow && \frac{-z+2}{z+1} &= \sqrt[3]{2} \\ \Rightarrow && -z+2 &= \sqrt[3]{2} z + \sqrt[3]{2} \\ \Rightarrow && z &= \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1} \end{align*}
4. \(\,\) \begin{align*} && 0 &= x^3 - 3p^2x + 2p^3 \\ &&&= (x-p)(x^2+px-2p^2) \\ &&&=(x-p)^2(x+2p)\\ \Rightarrow && x &= p, p, -2p \end{align*} Therefore if we have a repeated root to our associated quadratic we can find a cubic of the form \(x^3-3p^2x+2p^3\), but we know this has roots we can find.

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Solution:

1. \(n=1\): \begin{align*} && p_1(x) &= (x+1)^2 - 3x(x^2+x+1)^0 \\ &&&= x^2+2x+1-3x \\ &&&= x^2-x+1\\ && q_1(x) &= \frac{x^3+1}{x+1} \\ &&&= x^2-x+1 = p_1(x) \\ \\ && p_2(x) &= (x+1)^4-5x(x^2+x+1)^1 \\ &&&= x^4+4x^3+6x^2+4x+1 - 5x^3-5x^2-5x \\ &&&= x^4-x^3+x^2-x+1 \\ &&q_2(x) &= \frac{x^5+1}{x+1} \\ &&&= x^4-x^3+x^2-x+1 = p_2(x) \\ \\ && p_3(x) &= (x+1)^6-7x(x^2+x+1)^2 \\ &&&= x^6+6x^5+15x^4+20x^3+15x^2+6x+1 - 7x(x^4+2x^3+3x^2+2x+1) \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 \\ && q_3(x) &= \frac{x^7+1}{x+1} \\ &&&= x^6-x^5+x^4-x^3+x^2-x+1 = p_3(x) \\ \\ && p_4(1) &= 2^8 - 9 \cdot 1 \cdot 3^3 \\ &&&= 256 - 243 = 13 \\ && q_4(1) &= \frac{2}{2} = 1 \neq 13 \end{align*}
2. • \(\bf (a)\) \(\,\) \begin{align*} && \frac{300^3+1}{300+1} &= (300+1)^2 - 3 \cdot 300 \\ &&&= 301^2 - 30^2 \\ &&&= 271 \cdot 331 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && \dfrac {7^{49}+1}{7^7+1} &= (7^7+1)^6 - 7 \cdot 7^7 \cdot (7^2+7+1)^2 \\ &&&= (7^7+1)^6 - 7^8 \cdot (7^2+7+1)^2 \\ &&&= ((7^7+1)^3 - 7^4(7^2+7+1)) \cdot ((7^7+1)^3 + 7^4(7^2+7+1)) \end{align*}

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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. \(\,\) \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 \(4 \times 3 = 12\)

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Solution:

1. \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
   

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2. \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
   

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3. If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is: \begin{align*} 0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\ &= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y \end{align*} which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\). However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
   

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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 \(6^4 = 8^3 + 28^2\)

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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:

1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

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Solution: \begin{align*} && x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\ &&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\ \end{align*} We must have: \begin{align*} && 0 &= ak^2 + bk+c \tag{1}\\ &&0 &= bk^2+ck+a \tag{2}\\ b*(1)&& 0 &= abk^2 + b^2k+cb \\ a*(2)&& 0 &= abk^2 + ack + a^2 \\ \Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\ \Rightarrow && (ac-b^2)k &= bc-a^2 \\ \\ c*(1) && 0 &= ack^2+bck+c^2 \\ b*(2) && 0 &= b^2k^2+bck+ab \\ \Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\ \Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\ \\ \Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\ \Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\ \Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\ \Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\ &&&= a(a^3+b^3+c^3-3abc) \\ \underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\ &&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align*} Therefore \(a+b+c = 0\). (Since otherwise \(a=b=c\) but \(ac \neq b^2\)). This means \(1\) is a root of our equations. Therefore, either \(k = 1\) or they have both roots in common, ie they are the same equation up to a scalar factor. ie \(b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1\). Therefore, they are the same equation.

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Solution:

1. \(\,\) \begin{align*} && x^2-y^2 &=(x-y)^3 \\ \Rightarrow && x+y &=d^2 \\ && x-y &= d \\ \Rightarrow && x &= \tfrac12(d^2+d) \\ && y &= \tfrac12(d^2-d) \end{align*} Therefore consider \(x^2 = m, y^2 = n\), so \(m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2\) so we want \(d^2-d > 20\), so \(d = 6, n = 225, m = 441\).
2. \(\,\) \begin{align*} && x^3-y^3 &= (x-y)^4 \\ \Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\ && d^3 &= (x-y)^2+3xy \\ && d^3 &= d^2 + 3xy \\ \Rightarrow && 3xy &= d^3 - d^2 \\ \Rightarrow && 3x(x-d) &= d^3-d^2 \\ \Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\ \Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\ &&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\ &&&= d \pm d \sqrt{\frac{4d-1}{3}} \end{align*} Therefore we need \(\frac{4d-1}{3}\) to be an odd square. \(y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}\). Since we want positive values, we should take the positive square roots. \(d = \frac{3 \cdot 3^2 + 1}{4} = 7\) we have \(2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7\)

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Solution: \begin{align*} && \alpha \sin(A-B) + \beta \cos (A + B) &= \gamma \sin(A+B) \\ \Leftrightarrow && \alpha \sin A \cos B - \alpha \cos A \sin B + \beta \cos A \cos B - \beta \sin A \sin B &= \gamma \sin A \cos B + \gamma \cos A \sin B \\ \Leftrightarrow && \alpha \tan A - \alpha \tan B + \beta - \beta \tan A \tan B &= \gamma \tan A + \gamma \tan B \\ \Leftrightarrow && \beta \tan A \tan B +(\gamma-\alpha) \tan A + (\gamma +\alpha)\tan B&=\beta \\ \Leftrightarrow && \tan A \tan B +\left (\frac{\gamma-\alpha}{\beta} \right) \tan A + \left (\frac{\gamma +\alpha}{\beta} \right)\tan B&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) - \frac{\gamma^2 - \alpha^2}{\beta^2}&=1\\ \Leftrightarrow && \left ( \tan A + \left ( \frac{\gamma+\alpha}{\beta} \right) \right) \left ( \tan B + \left ( \frac{\gamma-\alpha}{\beta} \right)\right) &= \frac{\beta^2+\gamma^2-\alpha^2}{\beta^2}\\ \end{align*} Which has the desired form iff \(\beta^2+\gamma^2 = \alpha^2\).

1. \(\,\) \begin{align*} && 2\sin(x-\tfrac14\pi) + \sqrt 3 \cos(x+\tfrac14\pi) &=\sin(x+\tfrac14\pi) \\ && 3 + 1 &= 4 \\ \Rightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{4} + \frac{1-2}{\sqrt3}\right) &= 0\\ \Rightarrow && \tan x &= -\sqrt3 \\ \Rightarrow && x &= \tfrac23\pi, \tfrac53\pi \end{align*}
2. \(\,\) \begin{align*} && 2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) &=\sin(x+\frac16\pi) \\ \Leftrightarrow && \left (\tan x + \frac{1+2}{\sqrt3} \right) \left ( \tan \frac{\pi}{3} + \frac{1-2}{\sqrt3}\right) &= 0\\ && x &\in [0, 2\pi) \end{align*}
3. \(\,\) \begin{align*} && 2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x) \\ && A-B =x + \tfrac13\pi, A+B &= 3x \\ \Rightarrow && A = 2x + \tfrac\pi6, B &= x-\tfrac{\pi}{6} \\ \Rightarrow && \tan (2x+\tfrac\pi6)&=-\sqrt3 \\ && 2x + \tfrac{\pi}{6} &= \tfrac23\pi, \tfrac53\pi, \tfrac83 \pi, \tfrac{11}3\pi \\ && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \\ && \tan(-x-\tfrac{\pi}{6}) &= \frac1{\sqrt{3}} \\ \Rightarrow && x-\tfrac{\pi}{6} &= \ldots, \tfrac{\pi}{6}, \tfrac{7\pi}{6}, \ldots \\ \Rightarrow && x &= \tfrac{\pi}3, \tfrac{4\pi}{3} \\ \\ \Rightarrow && x &= \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} , \tfrac{\pi}3, \tfrac{4\pi}{3} \end{align*}

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Solution: \begin{align*} 180^2 &= 32400 \\ 181^2 &= 32761 \\ 182^2 &= 33124 \\ 183^2 &= 33489 \\ 184^2 &= 33856 \end{align*} Therefore \(182^2 < 33\,127 < (182+1)^2\). and \((182+2)^2 - 33\,127 = 729 = 27^2\). Therefore \(33\,127 = 184^2 - 27^2 = 211 \times 157\). (Note both of these numbers are prime). Suppose \((n+m)^2 - 33\,127 = k^2\) then \(33\,127 = (n+m)^2-k^2 = (n+m-k)(n+m+k)\). Since there are only two factorisations of \(33\,127\) into positive integer factors with one factor larger than the other, the other factorisation must be: \(n+m+k = 33\,127, n+m-k = 1 \Rightarrow k = \frac{33\, 126}{2} = 16563\), ie \(16564^2 - 33\,127 = 16563^2\)