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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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2011 Paper 3 Q7
D: 1700.0 B: 1486.2
proof by induction surds binomial expansion integer expressions Pell equation algebraic identity nested radicals consecutive integers

Let \[ T _n = \left( \sqrt{a+1} + \sqrt a\right)^n\,, \] where \(n\) is a positive integer and \(a\) is any given positive integer.

  1. In the case when \(n\) is even, show by induction that \(T_n\) can be written in the form \[ A_n +B_n \sqrt{a(a+1)}\,, \] where \(A_n\) and \(B_n\) are integers (depending on \(a\) and \(n\)) and \(A_n^2 =a(a+1)B_n^2 +1\).
  2. In the case when \(n\) is odd, show by considering \((\sqrt{a+1} +\sqrt a)T_m\) where \(m\) is even, or otherwise, that \(T_n\) can be written in the form \[ C_n \sqrt {a+1} + D_n \sqrt a \,, \] where \(C_n\) and \(D_n\) are integers (depending on \(a\) and \(n\)) and \( (a+1)C_n^2 = a D_n^2 +1\,\).
  3. Deduce that, for each \(n\), \(T_n\) can be written as the sum of the square roots of two consecutive integers.

Solution:

  1. Claim: For all \(n \geq 1\) \(T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}\) where \(A_{2n}, B_{2n}\) are integers and \(A_{2n}^2 = a(a+1)B_{2n}^2+1\) Proof: (By induction) Base case: \(n =1\). \begin{align*} && T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= a+1+2\sqrt{a(a+1)}+a \\ &&&=2a+1+2\sqrt{a(a+1)} \\ \Rightarrow && A_2 &= 2a+1 \\ && B_2 &= 2 \\ && A_2^2 &= 4a^2+4a+1 \\ && a(a+1)B_1^2 + 1 &= 4a^2+4a+1 \end{align*} Therefore our base case is true. Suppose it is true for some \(n = k\) then consider \(n = k+1\) we must have \(T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}\) \begin{align*} && T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\ &&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\ \Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\ && B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\ && A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\ &&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\ &&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\ &&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\ && a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\ &&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\ &&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\ &&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1 \end{align*} So our relation holds ad therefore by induction we're done.
  2. When \(n\) is odd, notice that \begin{align*} && T_n &= (\sqrt{a+1}+\sqrt{a})(A_m+B_m\sqrt{a(a+1)})\\ &&&= \sqrt{a+1}(\underbrace{A_m+aB_m}_{C_n})+\sqrt{a}(\underbrace{(a+1)B_m+A_m}_{D_n}) \\ \\ && (a+1)C_n^2 &= (a+1)(A_m+aB_m)^2 \\ &&&= (a+1)(A_m^2+2aA_mB_m+B_m^2) \\ &&&= (a+1)(a(a+1)B_m^2+1+2aA_mB_m+B_m^2) \\ &&&= a(a+1)^2B_m^2 + 2a(a+1)A_mB_m+(a+1)B_m^2+a+1 \\ && aD_n^2+1 &= a((a+1)^2B_m + 2(a+1)A_mB_m + A_m^2)+1 \\ &&&= a(a+1)^2B_m + 2a(a+1)A_mB_m + aB_{m}^2+a+1 \end{align*}
  3. For even \(n\) \(T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}\) For odd \(n\), \(T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}\) therefore it is always the sum of the square root of two consecutive integers.

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