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2023 Paper 3 Q3
D: 1500.0 B: 1500.0
complex numbers argand diagram isosceles triangle roots of polynomials curve sketching cubic equations inequality

  1. Show that, if \(a\) and \(b\) are complex numbers, with \(b \neq 0\), and \(s\) is a positive real number, then the points in the Argand diagram representing the complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\) form an isosceles triangle. Given three points which form an isosceles triangle in the Argand diagram, explain with the aid of a diagram how to determine the values of \(a\), \(b\) and \(s\) so that the vertices of the triangle represent complex numbers \(a + sbi\), \(a - sbi\) and \(a + b\).
  2. Show that, if the roots of the equation \(z^3 + pz + q = 0\), where \(p\) and \(q\) are complex numbers, are represented in the Argand diagram by the vertices of an isosceles triangle, then there is a non-zero real number \(s\) such that \[\frac{p^3}{q^2} = \frac{27(3s^2 - 1)^3}{4(9s^2 + 1)^2}\,.\]
  3. Sketch the graph \(y = \dfrac{(3x-1)^3}{(9x+1)^2}\), identifying any stationary points.
  4. Show that if the roots of the equation \(z^3 + pz + q = 0\) are represented in the Argand diagram by the vertices of an isosceles triangle then \(\dfrac{p^3}{q^2}\) is a real number and \(\dfrac{p^3}{q^2} > -\dfrac{27}{4}\).

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2011 Paper 2 Q2
D: 1600.0 B: 1516.0
proof number theory cubic equations Diophantine equation perfect square inequality algebraic manipulation integer solutions

Write down the cubes of the integers \(1, 2, \ldots , 10\). The positive integers \(x\), \(y\) and \(z\), where \(x < y\), satisfy \[ x^3+y^3 = kz^3\,, \tag{\(*\)} \] where \(k\) is a given positive integer.

  1. In the case \(x+y =k\), show that \[ z^3 = k^2 -3kx+3x^2\,. \] Deduce that \((4z^3 - k^2)/3\) is a perfect square and that \(\frac14 {k^2} \le z^3 < k^2\,\). Use these results to find a solution of \((*)\) when \(k=20\).
  2. By considering the case \(x+y = z^2\), find two solutions of \((*)\) when \(k=19\).

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 \(3^3+5^3=19 \cdot 2^3\)

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2010 Paper 3 Q4
D: 1700.0 B: 1516.0
polynomials common roots quadratic equations cubic equations resultant proof algebraic manipulation

  1. The number \(\alpha\) is a common root of the equations \(x^2 +ax +b=0\) and \(x^2+cx+d=0\) (that is, \(\alpha\) satisfies both equations). Given that \(a\ne c\), show that \[ \alpha =- \frac{b-d}{a-c}\,. \] Hence, or otherwise, show that the equations have at least one common root if and only if \[ (b-d)^2 -a(b-d)(a-c) + b(a-c)^2 =0\,. \] Does this result still hold if the condition \(a\ne c\) is not imposed?
  2. Show that the equations \(x^2+ax+b=0\) and \(x^3+(a+1)x^2+qx+r=0\) have at least one common root if and only if \[ (b-r)^2-a(b-r)(a+b-q) +b(a+b-q)^2=0\,. \] Hence, or otherwise, find the values of \(b\) for which the equations \(2x^2+5 x+2 b=0\) and \(2x^3+7x^2+5x+1=0\) have at least one common root.

Solution:

  1. \begin{align*} && 0 &= \alpha^2 + a \alpha + b \tag{1} \\ && 0 &= \alpha^2 + c \alpha + d \tag{2} \\ \\ (1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\ \Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)} \end{align*} (\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have: \begin{align*} && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \end{align*} (\(\Leftarrow\)) Suppose the equation holds, then \begin{align*} && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\ \Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \end{align*} So \(\alpha\) is a root of the first equation. Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations. If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
  2. \begin{align*} && 0 &= x^2+ax+b \tag{1} \\ && 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\ \\ (2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3} \end{align*} Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition. \(a = \frac52, q = \frac52, r = \frac12\) \begin{align*} && 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\ &&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\ &&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\ &&&= (b-1)(4b^2-2b-1) \\ \Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}

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2003 Paper 1 Q7
D: 1484.0 B: 1516.0
proof inequality divisibility digit manipulation number theory cubic equations systematic search integer constraints

Let \(k\) be an integer satisfying \(0\le k \le 9\,\). Show that \(0\le 10k-k^2\le 25\) and that \(k(k-1)(k+1)\) is divisible by \(3\,\). For each \(3\)-digit number \(N\), where \(N\ge100\), let \(S\) be the sum of the hundreds digit, the square of the tens digit and the cube of the units digit. Find the numbers \(N\) such that \(S=N\). [Hint: write \(N=100a+10b+c\,\) where \(a\,\), \(b\,\) and \(c\) are the digits of \(N\,\).]

Solution: First note that \(10k - k^2 = 25 - (5-k)^2\) which is clearly bounded above by \(25\). The smallest it can be is when \(|5-k|\) is as large as possible, ie when \(k =0\) and we get a lower bound of \(0\). For \((k-1)k(k+1)\) notice this is the product of \(3\) consecutive integers, and therefore must be divisible by \(3\). (In fact, it's divisible by six, since \(\binom{k+1}{3}\) is the number of ways to choose \(3\) objects from \(k+1\). Let \(N = 100a + 10b + c\) where \(0 \leq a,b,c < 10\) and \(1 \leq a\). \(S = a + b^2 + c^3\) we want to find \begin{align*} && 100a +10b + c &= a + b^2 + c^3 \\ \Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\ \end{align*} Therefore \(3 \mid 10b - b^2 = b(10-b)\). Therefore \(3 \mid b\) or \(3 \mid 10-b\) so \(b = 0, 3, 6, 1, 4, 7\) We also have \(99a \geq 99\) and \(10b-b^2 \in [0, 25]\) so we need \(c^3-c \geq 99\), so \(c \geq 5\) Case \(c = 5\), Then \(c^3-c = 120\) so \(a = 1\) and \(10b-b^2 = 21\) so \(b= 3, 7\) \(N = 135, 175\) Case \(c = 6\), so \(c^3 - c = 210\) so \(a = 2\) and \(25-(5-k)^2 = 12\) so no solutions. Case \(c = 7\), so \(7^3 - 7 = 336\) so \(a = 3\) and \(25-(5-k)^2 = 39\) so no solutions. Case \(c = 8\) so \(8^3-8 = 504\) so \(a = 5\) and \(25-(5-k)^2 = 9\), so \(b = 1, 9\) and \(N = 518, 598\) Case \(c = 9\) so \(9^3 - 9 = 720\), so \(a = 7\) and \(25-(5-k)^2 = 27\) so no solutions. Therefore all the solutions are \(N = 135, 175, 518, 598\)

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1998 Paper 2 Q1
D: 1600.0 B: 1500.0
proof proof by contradiction number theory divisibility parity cubic equations integer solutions Diophantine equation

Show that, if \(n\) is an integer such that $$(n-3)^3+n^3=(n+3)^3,\quad \quad {(*)}$$ then \(n\) is even and \(n^2\) is a factor of \(54\). Deduce that there is no integer \(n\) which satisfies the equation \((*)\). Show that, if \(n\) is an integer such that $$(n-6)^3+n^3=(n+6)^3, \quad \quad{(**)}$$ then \(n\) is even. Deduce that there is no integer \(n\) which satisfies the equation \((**)\).

Solution: \begin{align*} && n^3 &= (n+3)^3 - (n-3)^3 \\ &&&= n^3 + 9n^2+27n + 27 - (n^3 - 9n^2+27n-27) \\ &&&= 18n^2+54 \end{align*} Therefore since \(2 \mid 2(9n^2 + 27)\), \(2 \mid n^3 \Rightarrow 2 \mid n\), so \(n\) is even. Since \(n^2 \mid n^3\), \(n^2 \mid 54 = 2 \cdot 3^3\), therefore \(n = 1\) or \(n = 3\). \((1-3)^3 + 1^3 < 0 < (1+3)^3\). So \(n = 1\) doesn't work. \((3 - 3)^3 + 3^3 < (3+3)^3\) so \(n = 3\) doesn't work. Therefore there are no solutions. \begin{align*} && n^3 &= (n+6)^3 - (n-6)^3 \\ &&&= n^3 + 18n^2 + 180n + 6^3 - (n^3 - 18n^2 + 180n - 6^3 ) \\ &&&= 36n^2+2 \cdot 6^3 \end{align*} Therefore \(n^2 \mid 2 \cdot 6^3 = 2^4 \cdot 3^3\), therefore \(n = 1, 2, 3, 4, 6, 12\). \(n = 1\), \(1^3 <36+2\cdot 6^3\) \(n = 2\), \(2^3 <36 \cdot 4 + 2 \cdot 6^3\) \(n = 3\), \(3^3 <36 \cdot 9 + 2 \cdot 6^3\) \(n = 4\), \(4^3 < 36 \cdot 16 + 2 \cdot 6^3\) \(n = 6\), \(6^3 < 36\cdot 6^2+ 2 \cdot 6^3\) \(n = 12\), \(12^3 < 36 \cdot 12^2 + 2 \cdot 6^3\) Therefore there are no solutions \(n\) to the equation. These are both special cases of Fermat's Last Theorem, when \(n = 3\)

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