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2024 Paper 2 Q1
D: 1500.0 B: 1500.0
proof consecutive integers diophantine equation number theory algebraic manipulation integer solutions

In the equality \[ 4 + 5 + 6 + 7 + 8 = 9 + 10 + 11, \] the sum of the five consecutive integers from 4 upwards is equal to the sum of the next three consecutive integers. Throughout this question, the variables \(n\), \(k\) and \(c\) represent positive integers.

  1. Show that the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers if and only if \[ 2n^2 + k = 2ck + k^2. \]
  2. Find the set of possible values of \(n\), and the corresponding values of \(c\), in each of the cases
    1. \(k = 1\)
    2. \(k = 2\).
  3. Show that there are no solutions for \(c\) and \(n\) if \(k = 4\).
  4. Consider now the case where \(c = 1\).
    1. Find two possible values of \(k\) and the corresponding values of \(n\).
    2. Show, given a possible value \(N\) of \(n\), and the corresponding value \(K\) of \(k\), that \[ N' = 3N + 2K + 1 \] will also be a possible value of \(n\), with \[ K' = 4N + 3K + 1 \] as the corresponding value of \(k\).
    3. Find two further possible values of \(k\) and the corresponding values of \(n\).

Solution:

  1. Suppose the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers then \begin{align*} && \sum_{i=c}^{i=c+n+k-1} i &= \sum_{i=c+n+k}^{c+2n+k-1} i \\ \Leftrightarrow && \frac{(c+n+k-1)(c+n+k)}{2} - \frac{(c-1)c}{2} &= \frac{(c+2n+k-1)(c+2n+k)}{2} - \frac{(c+n+k-1)(c+n+k)}{2} \\ \Leftrightarrow && 2(c+n+k-1)(c+n+k) &= (c+2n+k-1)(c+2n+k) + c(c-1) \\ \Leftrightarrow && 2c^2+4cn+4ck+2n^2+4kn+2k^2-2c-2n-2k&=2c^2+4cn+2ck+4n^2+4nk+k^2-2c-2n-k \\ \Leftrightarrow && 2ck+k^2&=2n^2+k \\ \end{align*}
    1. If \(k=1\) then \begin{align*} && 2n^2 + 1 &= 2c + 1 \\ \Rightarrow && c &= n^2 \end{align*} So \(n\) can take any value and \(c = n^2\)
    2. If \(k=2\) then \begin{align*} && 2n^2+2&= 4c+4 \\ \Rightarrow && n^2-1 &=2c \end{align*} So \(n\) must be odd, and \(c = \frac12(n^2-1)\)
  3. Suppose \(k=4\) then \(2n^2+4 = 8c+16\) or \(n^2-6 = 4c\) but then the left hand side is \(2, 3 \pmod{4}\) which is a contradiction.
  4. Suppose \(c =1\)
    1. Since \(2n^2+k = 2k + k^2\) or \(2n^2 = k^2+k\) we can have \(k = 1, n = 1\) or \(k = 8, n = 6\)
    2. Suppose \(2N^2 = K^2 + K\) then consider \begin{align*} && 2(N')^2 &= 2(3N+2K+1)^2 \\ &&&= 2(9N^2+4K^2+1+12NK+6N+4K) \\ &&&= 18N^2+8K^2+24NK+12N+8K+2 \\ && (K')^2+K' &= (4N+3K+1)^2 + (4N+3K+1) \\ &&&= 16N^2 + 9K^2+1+24NK+12N+9K+1 \\ &&&= 16N^2+9K^2+24NK+12N+9K+2 \\ \Rightarrow && 2(N')^2-(K')^2-K' &= 2N^2-K^2-K \\ &&&= 0 \end{align*} as required.
    3. So consider \((k,n) = (1,1), (8,6), (49, 35), (288,204)\)

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2024 Paper 2 Q5
D: 1500.0 B: 1500.0
polynomials completing the square range of function integer solutions quadratic equations inequality set intersection

  1. The functions \(\mathrm{f}_1\) and \(\mathrm{F}_1\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_1(n) = n^2 + 6n + 11, \] \[ \mathrm{F}_1(n) = n^2 + 2. \] Show that \(\mathrm{F}_1\) has the same range as \(\mathrm{f}_1\).
  2. The function \(\mathrm{g}_1\), with domain \(\mathbb{Z}\), is defined by \[ \mathrm{g}_1(n) = n^2 - 2n + 5. \] Show that the ranges of \(\mathrm{f}_1\) and \(\mathrm{g}_1\) have empty intersection.
  3. The functions \(\mathrm{f}_2\) and \(\mathrm{g}_2\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_2(n) = n^2 - 2n - 6, \] \[ \mathrm{g}_2(n) = n^2 - 4n + 2. \] Find any integers that lie in the intersection of the ranges of the two functions.
  4. Show that \(p^2 + pq + q^2 \geqslant 0\) for all real \(p\) and \(q\). The functions \(\mathrm{f}_3\) and \(\mathrm{g}_3\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_3(n) = n^3 - 3n^2 + 7n, \] \[ \mathrm{g}_3(n) = n^3 + 4n - 6. \] Find any integers that lie in the intersection of the ranges of the two functions.

Solution:

  1. \(\,\) \begin{align*} && f_1(n) &= n^2 + 6n + 11 \\ &&&= (n+3)^2 + 2 \\ &&&=F_1(n+3) \end{align*} Since \(n \mapsto n+3\) is a bijection on \(\mathbb{Z}\) both functions must have exactly the same range.
  2. \(g_1(n) = n^2-2n+5 = (n-1)^2 + 4\). Since squares are always \(0, 1 \pmod{4}\) it's impossible for \(f_1\) and \(g_1\) to take the same value therefore the ranges have empty intersection.
  3. \(\,\) \begin{align*} && f_2(n) &= n^2-2n - 6 \\ &&&= (n-1)^2-7 \\ && g_2(n) &= n^2-4n+2 \\ &&&= (n-2)^2 - 2 \end{align*} so suppose \(x^2 - 7 = y^2 - 2\) then \begin{align*} && x^2 - 7 &= y^2 -2 \\ \Rightarrow && 5 &= y^2 - x^2 \\ &&&= (y-x)(y+x) \end{align*} So we have cases: \(y-x = -5, y + x = -1 \Rightarrow y = -3\) and the output is \(7\) \(y-x=-1, y+x = -5 \Rightarrow y = -3\) same output \(y-x=1, y+x = 5 \Rightarrow y = 3\) same output \(y-x=5, y-x = 1 \Rightarrow y = 3\) same ouput.
  4. \begin{align*} && 0 &\leq \frac12(p^2+q^2)+\frac12(p+q)^2 \\ &&&= p^2 + q^2 + pq \end{align*} Looking at \(f_3\) we see \begin{align*} && f_3(n) &= n^3 - 3n^2 + 7n \\ &&&= (n-1)^3 -3n + 7n +1 \\ &&&= (n-1)^3 +4(n-1) -3 \\ &&&= g_3(n-1) + 3 \end{align*} So suppose we have two values which are equal, ie \begin{align*} && x^3 + 4x -3 &= y^3 +4y -6 \\ \Rightarrow && 3 &= y^3-x^3+4y-4x \\ &&&= (y-x)(y^2+xy+x^2+4) \end{align*} Since \(x^2+xy+y^2 \geq 0\) then the right hand factor is always a positive integer bigger than \(3\) and in particular there will be no solutions and hence no integers in the intersection of the ranges.

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2023 Paper 2 Q7
D: 1500.0 B: 1500.0
complex numbers modulus sum of two squares number theory proof algebraic identity integer solutions

  1. The complex numbers \(z\) and \(w\) have real and imaginary parts given by \(z = a + \mathrm{i}b\) and \(w = c + \mathrm{i}d\). Prove that \(|zw| = |z||w|\).
  2. By considering the complex numbers \(2 + \mathrm{i}\) and \(10 + 11\mathrm{i}\), find positive integers \(h\) and \(k\) such that \(h^2 + k^2 = 5 \times 221\).
  3. Find positive integers \(m\) and \(n\) such that \(m^2 + n^2 = 8045\).
  4. You are given that \(102^2 + 201^2 = 50805\). Find positive integers \(p\) and \(q\) such that \(p^2 + q^2 = 36 \times 50805\).
  5. Find three distinct pairs of positive integers \(r\) and \(s\) such that \(r^2 + s^2 = 25 \times 1002082\) and \(r < s\).
  6. You are given that \(109 \times 9193 = 1002037\). Find positive integers \(t\) and \(u\) such that \(t^2 + u^2 = 9193\).

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2023 Paper 3 Q5
D: 1500.0 B: 1500.0
number theory diophantine equations factorisation prime numbers proof by contradiction integer solutions

  1. Show that if \[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,,\] then \((2x - 7)(y - 7) = 49\). By considering the factors of \(49\), find all the pairs of positive integers \(x\) and \(y\) such that \[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,.\]
  2. Let \(p\) and \(q\) be prime numbers such that \[p^2 + pq + q^2 = n^2\] where \(n\) is a positive integer. Show that \[(p + q + n)(p + q - n) = pq\] and hence explain why \(p + q = n + 1\). Hence find the possible values of \(p\) and \(q\).
  3. Let \(p\) and \(q\) be positive and \[p^3 + q^3 + 3pq^2 = n^3\,.\] Show that \(p + q - n < p\) and \(p + q - n < q\). Show that there are no prime numbers \(p\) and \(q\) such that \(p^3 + q^3 + 3pq^2\) is the cube of an integer.

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2019 Paper 3 Q4
D: 1500.0 B: 1500.0
roots of polynomials symmetric functions Vieta's formulas Newton's sums polynomial proof by contradiction integer solutions classification

The \(n\)th degree polynomial P\((x)\) is said to be reflexive if:

  1. P\((x)\) is of the form \(x^n - a_1x^{n-1} + a_2x^{n-2} - \cdots + (-1)^na_n\) where \(n \geq 1\);
  2. \(a_1, a_2, \ldots, a_n\) are real;
  3. the \(n\) (not necessarily distinct) roots of the equation P\((x) = 0\) are \(a_1, a_2, \ldots, a_n\).
  1. Find all reflexive polynomials of degree less than or equal to 3.
  2. For a reflexive polynomial with \(n > 3\), show that $$2a_2 = -a_2^2 - a_3^2 - \cdots - a_n^2.$$ Deduce that, if all the coefficients of a reflexive polynomial of degree \(n\) are integers and \(a_n \neq 0\), then \(n \leq 3\).
  3. Determine all reflexive polynomials with integer coefficients.

Solution:

  1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
  2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
  3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

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2013 Paper 2 Q7
D: 1600.0 B: 1516.0
proof Pell equation Diophantine equation integer solutions number theory factorisation triangular numbers prime factorisation

  1. Write down a solution of the equation \[ x^2-2y^2 =1\,, \tag{\(*\)} \] for which \(x\) and \(y\) are non-negative integers. Show that, if \(x=p\), \(y=q\) is a solution of (\(*\)), then so also is \(x=3p+4q\), \(y=2p+3q\). Hence find two solutions of \((*)\) for which \(x\) is a positive odd integer and \(y\) is a positive even integer.
  2. Show that, if \(x\) is an odd integer and \(y\) is an even integer, \((*)\) can be written in the form \[ n^2 = \tfrac12 m(m+1)\,, \] where \(m\) and \(n\) are integers.
  3. The positive integers \(a\), \(b\) and \(c\) satisfy \[ b^3=c^4-a^2\,, \] where \(b\) is a prime number. Express \(a\) and \(c^2\) in terms of \(b\) in the two cases that arise. Find a solution of \(a^2+b^3=c^4\), where \(a\), \(b\) and \(c\) are positive integers but \(b\) is not prime.

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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2011 Paper 1 Q8
D: 1516.0 B: 1484.0
proof integer solutions Diophantine equation cubic inequality number theory squeezing argument

  1. The numbers \(m\) and \(n\) satisfy \[ m^3=n^3+n^2+1\,. \tag{\(*\)} \]
    • Show that \(m > n\). Show also that \(m < n+1\) if and only if \(2n^2+3n > 0\,\). Deduce that \(n < m < n+1\) unless \(-\frac32 \le n \le 0\,\).
    • Hence show that the only solutions of \((*)\) for which both \(m\) and \(n\) are integers are \((m,n) = (1,0)\) and \((m,n)= (1,-1)\).
  2. Find all integer solutions of the equation \[ p^3=q^3+2q^2-1\,. \]

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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 1 Q8
D: 1500.0 B: 1484.0
proof by contradiction divisibility infinite descent number theory modular arithmetic integer solutions proof

  1. Suppose that \(a\), \(b\) and \(c\) are integers that satisfy the equation \[ a^{3}+3b^{3}=9c^{3}. \] Explain why \(a\) must be divisible by 3, and show further that both \(b\) and \(c\) must also be divisible by 3. Hence show that the only integer solution is \(a=b=c=0\,\).
  2. Suppose that \(p\), \(q\) and \(r\) are integers that satisfy the equation \[ p^4 +2q^4 = 5r^4 \,.\] By considering the possible final digit of each term, or otherwise, show that \(p\) and \(q\) are divisible by 5. Hence show that the only integer solution is \(p=q=r=0\,\).

Solution:

  1. Since \(a^3 = 9c^3 - 3b^3 = 3(3c^3-b^3)\) we must have \(3 \mid a^3\). But since \(3\) is prime, \(3 \mid a\). Since \(3 \mid a\) we can write \(a = 3a'\) for some \(a' \in \mathbb{Z}\). Therefore our equation is \(27(a')^3 + 3b^3 = 9c^3 \Rightarrow 9(a')^3 + b^3 = 3c^3\) which means that \(3 \mid b\) by the same argument from earlier. So \(b = 3b'\) so the equation is \(9(a')^3 + 27(b')^3 = 3c^3 \Rightarrow 3(a')^3 + 9(b')^3 = c^3\) which means that \(3 \mid c\). Suppose \((a,b,c)\) is the smallest measured by \(a^2+b^2+c^2\) with \(a, b, c\neq 0\). Then \((\frac{a}{3}, \frac{b}{3}, \frac{c}{3})\) is also a solution. But this contradicts that we had found the smallest solution. Therefore the only possible solution is \((0,0,0)\) which clearly works.
  2. Consider \(p, q \pmod{5}\). By \(FLT\) \(p^4, q^4 = 0, 1 \pmod{5}\) so \(p^4+2q^4 \in \{0, 1, 2, 3\}\) and in particular the only way they are divisible by \(5\) is if \(p \equiv q \equiv 0 \pmod{5}\). Therefore \(p = 5p', q = 5q'\) and so \(5^4(p')^4 + 5^4(q')^4 = 5r^4 \Rightarrow r^4 = 5(25(p')^4 + 25(q')^4) \Rightarrow 5\mid r^4 \Rightarrow 5 \mid r\). Therefore we can use the same argument about the smallest solution to show that \(p = q= r = 0\)

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2007 Paper 1 Q2
D: 1500.0 B: 1515.7
trigonometry arctan tan addition formula integer solutions Diophantine equation highest common factor proof

  1. Given that \(A = \arctan \frac12\) and that \(B = \arctan\frac13\,\) (where \(A\) and \(B\) are acute) show, by considering \(\tan \left( A + B \right)\), that \(A + B = {\frac{1}{4}\pi }\). The non-zero integers \(p\) and \(q\) satisfy \[ \displaystyle \arctan {\frac1 p} + \arctan {\frac1 q} = {\frac\pi 4}\,. \] Show that \( \left ( p-1 \right) \left(q-1 \right) = 2\) and hence determine \(p\) and \(q\).
  2. Let \(r\), \(s\) and \(t\) be positive integers such that the highest common factor of \(s\) and \(t\) is \(1\). Show that, if \[ \arctan {\frac1 r} + \arctan \frac s {s+t} = {\frac\pi 4}\,, \] then there are only two possible values for \(t\), and give \(r\) in terms of \(s\) in each case.

Solution:

  1. \begin{align*} && \tan (A+B) &= \frac{\tan A + \tan B}{1-\tan A \tan B}\\ &&&= \frac{\tan \arctan \frac12 + \tan \arctan \frac13}{1-\tan \arctan \frac12 \tan \arctan \frac13}\\ &&&= \frac{\frac12+\frac13}{1-\frac16} \\ &&&= \frac{3+2}{5} \\ &&&= 1 \\ \Rightarrow && A+B &= \frac{\pi}{4} + n \pi \end{align*} but since \(A,B\) are acute \(0 < A+B < \pi\), so \(A+B = \frac{\pi}{4}\) \begin{align*} && 1 &= \tan \frac{\pi}{4} \\ &&&= \tan \left ( \arctan {\frac1 p} + \arctan {\frac1 q}\right) \\ &&&= \frac{\frac1p + \frac1q}{1-\frac1{pq}} \\ &&&= \frac{q+p}{pq-1} \\ \Rightarrow && pq-1 &= q+p \\ \Rightarrow && 0 &= pq-q-p-q \\ &&&= (p-1)(q-1)-2 \\ \Rightarrow && 2 &= (p-1)(q-1) \end{align*} But \(p\),\(q\) are integers, so \(p-1 \in \{-2,-1,1,1\} \Rightarrow p \in \{-1,0,2,3\}\) but we cannot have \(p= 0\), so we must have \((p,q) = (2,3), (3,2)\)
  2. \begin{align*} && 1 &= \tan \frac{\pi}{4} \\ &&&= \tan \left ( \arctan {\frac1 r} + \arctan \frac s {s+t} \right) \\ &&&= \frac{\frac1r + \frac{s}{s+t}}{1-\frac{s}{r(s+t)}} \\ &&&= \frac{s+t+sr}{r(s+t)-s} \\ \Rightarrow && rs+rt-s &= s+t + sr \\ \Rightarrow && 0 &= rt-2s-t \\ &&2s&= t(r-1) \end{align*} Since \((s,t) =1\), we must have \(t \mid 2\), so \( t = 1,2\) and \(r = 2s+1\) or \(r=s+1\) respectively.

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2007 Paper 1 Q6
D: 1500.0 B: 1489.2
polynomials difference of powers factorisation algebraic manipulation integer solutions substitution quadratic formula number theory

  1. Given that \(x^2 - y^2 = \left( x - y \right)^3\) and that \(x-y = d\) (where \(d \neq 0\)), express each of \(x\) and \(y\) in terms of \(d\). Hence find a pair of integers \(m\) and \(n\) satisfying \(m-n = \left( \sqrt {m} - \sqrt{n} \right)^3\) where \(m > n > 100\).
  2. Given that \(x^3 - y^3 = \left( x - y \right)^4\) and that \(x-y = d\) (where \(d \neq 0\)), show that \(3xy = d^3 - d^2\). Hence show that \[ 2x = d \pm d \sqrt {\frac{4d-1 }{3}} \] and determine a pair of distinct positive integers \(m\) and \(n\) such that \(m^3 - n^3 = \left( m - n \right)^4\).

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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2006 Paper 1 Q6
D: 1500.0 B: 1606.3
proof algebraic verification Pell equation substitution Diophantine equation integer solutions quadratic forms

  1. Show that, if \(\l a \, , b\r\) is any point on the curve \(x^2 - 2y^2 = 1\), then \(\l 3a + 4b \, , 2a + 3b \r\,\) also lies on the curve.
  2. Determine the smallest positive integers \(M\) and \(N\) such that, if \(\l a \,, b\r\) is any point on the curve \(Mx^2 - Ny^2 = 1\), then \((5a+6b\,, 4a+5b)\) also lies on the curve.
  3. Given that the point \(\l a \, , b\r\) lies on the curve \(x^2 - 3y^2 = 1\,\), find positive integers \(P\), \(Q\), \(R\) and \(S\) such that the point \((P a +Q b\,, R a + Sb)\) also lies on the curve.

Solution:

  1. Suppose \(a^2-2b^2=1\) then \begin{align*} (3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\ &=a^2-2b^2 \\ &= 1 \end{align*} Therefore \((3a+4b,2a+3b)\) also lies on the curve.
  2. Suppose \(Ma^2-Nb^2 = 1\) then \begin{align*} M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\ &= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2 \end{align*} Therefore we need \(3M = 2N\) so the smallest possible value would have to be \(M = 2, N = 3\), which does work
  3. Consider \(x + \sqrt{3}y\), then consider \((x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}\). Notice that \((x+\sqrt{3}y)(x-\sqrt{3}y) = 1\) and \((2+\sqrt{3})(2-\sqrt{3}) = 1\) so \(((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1\), so we can take \(P=2,Q=3,R=1,S=2\)

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2002 Paper 3 Q4
D: 1700.0 B: 1490.1
inequality proof by contradiction cubic quadratic integer solutions Diophantine equation algebraic manipulation number theory

Show that if \(x\) and \(y\) are positive and \(x^3 + x^2 = y^3 - y^2\) then \(x < y\,\). Show further that if \(0 < x \le y - 1\), then \(x^3 + x^2 < y^3 - y^2\). Prove that there does not exist a pair of {\sl positive} integers such that the difference of their cubes is equal to the sum of their squares. Find all the pairs of integers such that the difference of their cubes is equal to the sum of their squares.

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1998 Paper 1 Q12
D: 1484.0 B: 1606.9
probability independent events combinatorics divisibility integer solutions stars and bars complementary counting

Suppose that a solution \((X,Y,Z)\) of the equation \[X+Y+Z=20,\] with \(X\), \(Y\) and \(Z\) non-negative integers, is chosen at random (each such solution being equally likely). Are \(X\) and \(Y\) independent? Justify your answer. Show that the probability that \(X\) is divisible by \(5\) is \(5/21\). What is the probability that \(XYZ\) is divisible by 5?

Solution: They are not independent: \begin{align*} && \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\ && \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\ \end{align*} \begin{align*} X = 0: && 21 \text{ solutions} \\ X = 5: && 16 \text{ solutions} \\ X = 10: && 11 \text{ solutions} \\ X = 15: && 6 \text{ solutions} \\ X = 20: && 1 \text{ solutions} \\ 5 \mid X: && 55 \text{ solutions} \\ \\ && \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\ \Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21} \end{align*} \begin{align*} \mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\ &= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77} \end{align*}

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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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