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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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2022 Paper 3 Q2
D: 1500.0 B: 1500.0
number theory proof by contradiction infinite descent divisibility diophantine equation modular arithmetic

  1. Suppose that there are three non-zero integers \(a\), \(b\) and \(c\) for which \(a^3 + 2b^3 + 4c^3 = 0\). Explain why there must exist an integer \(p\), with \(|p| < |a|\), such that \(4p^3 + b^3 + 2c^3 = 0\), and show further that there must exist integers \(p\), \(q\) and \(r\), with \(|p| < |a|\), \(|q| < |b|\) and \(|r| < |c|\), such that \(p^3 + 2q^3 + 4r^3 = 0\). Deduce that no such integers \(a\), \(b\) and \(c\) can exist.
  2. Prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(9a^3 + 10b^3 + 6c^3 = 0\).
  3. By considering the expression \((3n \pm 1)^2\), prove that, unless an integer is a multiple of three, its square is one more than a multiple of \(3\). Deduce that the sum of the squares of two integers can only be a multiple of three if each of the integers is a multiple of three. Hence prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(a^2 + b^2 = 3c^2\).
  4. Prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(a^2 + b^2 + c^2 = 4abc\).

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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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2009 Paper 1 Q2
D: 1500.0 B: 1500.0
implicit differentiation tangent-normal cubic curve polynomial equation Diophantine equation algebraic manipulation

A curve has the equation \[ y^3 = x^3 +a^3+b^3\,, \] where \(a\) and \(b\) are positive constants. Show that the tangent to the curve at the point \((-a,b)\) is \[ b^2y-a^2x = a^3+b^3\,. \] In the case \(a=1\) and \(b=2\), show that the \(x\)-coordinates of the points where the tangent meets the curve satisfy \[ 7x^3 -3x^2 -27x-17 =0\,. \] Hence find positive integers \(p\), \(q\), \(r\) and \(s\) such that \[ p^3 = q^3 +r^3 +s^3\,. \]

Solution: \begin{align*} && y^3 &= x^3 + a^3 + b^3 \\ \Rightarrow && 3y^2 \frac{\d y}{\d x} &= 3x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x^2}{y^2} \end{align*} Therefore the tangent at the point \((-a,b)\) has gradient \(\frac{a^2}{b^2}\), ie \begin{align*} && \frac{y-b}{x+a} &= \frac{a^2}{b^2} \\ \Rightarrow && b^2y - b^3 &= a^2 x + a^3 \\ \Rightarrow && b^2 y-a^2 x &= a^3 + b^3 \end{align*} Notice that tangent will be, \(4y-x = 9\) so substituting this we obtain: \begin{align*} && \left (\frac{9+x}{4} \right)^3 &= x^3 + 9 \\ \Rightarrow && 9^3 + 3 \cdot 9^2 x + 3 \cdot 9x^2 + x^3 &= 64x^3 + 64 \cdot 9 \\ \Rightarrow && 9 \cdot (9^2 - 8^2) + 9 \cdot (3 \cdot 9) + 9 \cdot 3x^2 -9 \cdot 7x^3 &= 0 \\ \Rightarrow && 7x^3-3x^2-27x-17 &= 0 \\ \Rightarrow && (x+1)^2(7x-17) &= 0 \tag{repeated root since tangent} \end{align*} So we have another point on the curve \(y^3 = x^3 + 2^3 + 1^3\), namely \((\frac{17}7, \frac{17+9 \cdot 7}{28}) = (\frac{17}7, \frac{20}{7})\), so \begin{align*} 20^3 &= 17^3 + 14^3 + 7^3 \end{align*}

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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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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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2004 Paper 1 Q5
D: 1484.0 B: 1500.0
proof arithmetic progressions modular arithmetic divisibility proof by contradiction number theory Diophantine equation residues

The positive integers can be split into five distinct arithmetic progressions, as shown: \begin{align*} A&: \ \ 1, \ 6, \ 11, \ 16, \ ... \\ B&: \ \ 2, \ 7, \ 12, \ 17, \ ...\\ C&: \ \ 3, \ 8, \ 13, \ 18, \ ... \\ D&: \ \ 4, \ 9, \ 14, \ 19, \ ... \\ E&: \ \ 5, 10, \ 15, \ 20, \ ... \end{align*} Write down an expression for the value of the general term in each of the five progressions. Hence prove that the sum of any term in \(B\) and any term in \(C\) is a term in \(E\). Prove also that the square of every term in \(B\) is a term in \(D\). State and prove a similar claim about the square of every term in \(C\).

  1. Prove that there are no positive integers \(x\) and \(y\) such that \[ x^2+5y=243\,723 \,. \]
  2. Prove also that there are no positive integers \(x\) and \(y\) such that \[ x^4+2y^4=26\,081\,974 \,. \]

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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 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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1993 Paper 1 Q3
D: 1516.0 B: 1516.0
proof number theory Diophantine equation inequality integer solutions systematic case analysis unit fractions

  1. Find all the integer solutions with \(1\leqslant p\leqslant q\leqslant r\) of the equation \[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1\,, \] showing that there are no others.
  2. The integer solutions with \(1\leqslant p\leqslant q\leqslant r\) of \[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1\,, \] include \(p=1\), \(q=n,\) \(r=m\) where \(n\) and \(m\) are any integers satisfying \(1\leqslant m\leqslant n.\) Find all the other solutions, showing that you have found them all.

Solution:

  1. Suppose \(p > 3\) then there are clearly no solutions, since \(\frac1p+\frac1q+\frac1r \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} < 1\) Therefore there are 3 cases: \(p = 3 \Rightarrow p = q = r = 3\) \(p = 2\): \begin{align*} && \frac12 = \frac1q + \frac1r \\ \Rightarrow && 0 = qr - 2q-2q \\ \Rightarrow && 4 &= (q-2)(r-2) \\ \end{align*} Therefore \((p,q,r) = (2, 3, 6), (2, 4, 4)\) \(p = 1\) we have a contradiction the other way.
  2. We have already shown \(p < 3\), so we just need to check \(p = 2\) (since \(p=1\) is described in the question). \begin{align*} && \frac12 &< \frac1q+\frac1r \\ \Rightarrow && qr &< 2q+2r \\ \Rightarrow && 4 &> (q-2)(r-2) \\ \end{align*} Therefore we can have \((q-2)(r-2) = 0 \Rightarrow p = 2, q = 2, r = n\) Or we have have \((q-2)(r-2) = 1 \Rightarrow q = 3, r = 3\) Or we can have \((q-2)(r-2) = 2 \Rightarrow q = 3, r= 4\)

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1993 Paper 2 Q7
D: 1600.0 B: 1491.2
number theory modular arithmetic divisibility proof by contradiction infinite descent Diophantine equation mod 5

The integers \(a,b\) and \(c\) satisfy \[ 2a^{2}+b^{2}=5c^{2}. \] By considering the possible values of \(a\pmod5\) and \(b\pmod5\), show that \(a\) and \(b\) must both be divisible by \(5\). By considering how many times \(a,b\) and \(c\) can be divided by \(5\), show that the only solution is \(a=b=c=0.\)

Solution: \begin{array}{c|ccccc} a & 0 & 1 & 2 & 3 & 4 \\ a^2 & 0 & 1 & 4 & 4 & 1 \end{array} Therefore \(a^2 \in \{0,1,4\}\) and so we can have \begin{array} $2a^2+b^2 & 0 & 1 & 4 \\ \hline 0 & 0 & 1 & 4 \\ 1 & 2 & 3 & 1 \\ 4 & 3 & 4 & 2 \end{array} Therefore the only solution must have \(5 \mid a,b\), but then we can write them has \(5a'\) and \(5b'\) so the equation becomes \(2\cdot25 a'^2 + 25b'^2 = 5c^2\) ie \(5 \mid c^2 \Rightarrow 5 \mid c\). But that means we can always divide \((a,b,c)\) by \(5\), which is clearly a contradiction if we consider the lowest power of \(5\) dividing \(a,b,c\) for any solution.

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